Leetcode 题解 - 搜索

2022年5月21日大约 20 分钟

Leetcode 题解 - 搜索

Leetcode 题解 - 搜索

深度优先搜索和广度优先搜索广泛运用于树和图中，但是它们的应用远远不止如此。

BFS

广度优先搜索一层一层地进行遍历，每层遍历都是以上一层遍历的结果作为起点，遍历一个距离能访问到的所有节点。需要注意的是，遍历过的节点不能再次被遍历。

第一层：

0 -> {6,2,1,5}

第二层：

6 -> {4}
2 -> {}
1 -> {}
5 -> {3}

第三层：

4 -> {}
3 -> {}

每一层遍历的节点都与根节点距离相同。设 d_i 表示第 i 个节点与根节点的距离，推导出一个结论：对于先遍历的节点 i 与后遍历的节点 j，有 d_i <= d_j。利用这个结论，可以求解最短路径等 最优解 问题：第一次遍历到目的节点，其所经过的路径为最短路径。应该注意的是，使用 BFS 只能求解无权图的最短路径，无权图是指从一个节点到另一个节点的代价都记为 1。

在程序实现 BFS 时需要考虑以下问题：

队列：用来存储每一轮遍历得到的节点；
标记：对于遍历过的节点，应该将它标记，防止重复遍历。

1. 计算在网格中从原点到特定点的最短路径长度

1091. Shortest Path in Binary Matrix(Medium)

Leetcodeopen in new window / 力扣open in new window

[[1,1,0,1],
 [1,0,1,0],
 [1,1,1,1],
 [1,0,1,1]]

题目描述：0 表示可以经过某个位置，求解从左上角到右下角的最短路径长度。

public int shortestPathBinaryMatrix(int[][] grids) {
        if (grids == null || grids.length == 0 || grids[0].length == 0) {
            return -1;
        }
        int[][] direction = {{1, -1}, {1, 0}, {1, 1}, {0, -1}, {0, 1}, {-1, -1}, {-1, 0}, {-1, 1}};
        int m = grids.length, n = grids[0].length;
        Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
        queue.add(new Pair<>(0, 0));
        int pathLength = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            pathLength++;
            while (size-- > 0) {
                Pair<Integer, Integer> cur = queue.poll();
                int cr = cur.getKey(), cc = cur.getValue();
                if (grids[cr][cc] == 1) {
                    continue;
                }
                if (cr == m - 1 && cc == n - 1) {
                    return pathLength;
                }
                grids[cr][cc] = 1; // 标记
                for (int[] d : direction) {
                    int nr = cr + d[0], nc = cc + d[1];
                    if (nr < 0 || nr >= m || nc < 0 || nc >= n) {
                        continue;
                    }
                    queue.add(new Pair<>(nr, nc));
                }
            }
        }
        return -1;
    }

2. 组成整数的最小平方数数量

279. Perfect Squares (Medium)

Leetcodeopen in new window / 力扣open in new window

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

可以将每个整数看成图中的一个节点，如果两个整数之差为一个平方数，那么这两个整数所在的节点就有一条边。

要求解最小的平方数数量，就是求解从节点 n 到节点 0 的最短路径。

本题也可以用动态规划求解，在之后动态规划部分中会再次出现。

public int numSquares(int n) {
    List<Integer> squares = generateSquares(n);
    Queue<Integer> queue = new LinkedList<>();
    boolean[] marked = new boolean[n + 1];
    queue.add(n);
    marked[n] = true;
    int level = 0;
    while (!queue.isEmpty()) {
        int size = queue.size();
        level++;
        while (size-- > 0) {
            int cur = queue.poll();
            for (int s : squares) {
                int next = cur - s;
                if (next < 0) {
                    break;
                }
                if (next == 0) {
                    return level;
                }
                if (marked[next]) {
                    continue;
                }
                marked[next] = true;
                queue.add(next);
            }
        }
    }
    return n;
}

/**
 * 生成小于 n 的平方数序列
 * @return 1,4,9,...
 */
private List<Integer> generateSquares(int n) {
    List<Integer> squares = new ArrayList<>();
    int square = 1;
    int diff = 3;
    while (square <= n) {
        squares.add(square);
        square += diff;
        diff += 2;
    }
    return squares;
}

3. 最短单词路径

127. Word Ladder (Medium)

Leetcodeopen in new window / 力扣open in new window

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

题目描述：找出一条从 beginWord 到 endWord 的最短路径，每次移动规定为改变一个字符，并且改变之后的字符串必须在 wordList 中。

public int ladderLength(String beginWord, String endWord, List<String> wordList) {
    wordList.add(beginWord);
    int N = wordList.size();
    int start = N - 1;
    int end = 0;
    while (end < N && !wordList.get(end).equals(endWord)) {
        end++;
    }
    if (end == N) {
        return 0;
    }
    List<Integer>[] graphic = buildGraphic(wordList);
    return getShortestPath(graphic, start, end);
}

private List<Integer>[] buildGraphic(List<String> wordList) {
    int N = wordList.size();
    List<Integer>[] graphic = new List[N];
    for (int i = 0; i < N; i++) {
        graphic[i] = new ArrayList<>();
        for (int j = 0; j < N; j++) {
            if (isConnect(wordList.get(i), wordList.get(j))) {
                graphic[i].add(j);
            }
        }
    }
    return graphic;
}

private boolean isConnect(String s1, String s2) {
    int diffCnt = 0;
    for (int i = 0; i < s1.length() && diffCnt <= 1; i++) {
        if (s1.charAt(i) != s2.charAt(i)) {
            diffCnt++;
        }
    }
    return diffCnt == 1;
}

private int getShortestPath(List<Integer>[] graphic, int start, int end) {
    Queue<Integer> queue = new LinkedList<>();
    boolean[] marked = new boolean[graphic.length];
    queue.add(start);
    marked[start] = true;
    int path = 1;
    while (!queue.isEmpty()) {
        int size = queue.size();
        path++;
        while (size-- > 0) {
            int cur = queue.poll();
            for (int next : graphic[cur]) {
                if (next == end) {
                    return path;
                }
                if (marked[next]) {
                    continue;
                }
                marked[next] = true;
                queue.add(next);
            }
        }
    }
    return 0;
}

DFS

广度优先搜索一层一层遍历，每一层得到的所有新节点，要用队列存储起来以备下一层遍历的时候再遍历。

而深度优先搜索在得到一个新节点时立即对新节点进行遍历：从节点 0 出发开始遍历，得到到新节点 6 时，立马对新节点 6 进行遍历，得到新节点 4；如此反复以这种方式遍历新节点，直到没有新节点了，此时返回。返回到根节点 0 的情况是，继续对根节点 0 进行遍历，得到新节点 2，然后继续以上步骤。

从一个节点出发，使用 DFS 对一个图进行遍历时，能够遍历到的节点都是从初始节点可达的，DFS 常用来求解这种 可达性 问题。

在程序实现 DFS 时需要考虑以下问题：

栈：用栈来保存当前节点信息，当遍历新节点返回时能够继续遍历当前节点。可以使用递归栈。
标记：和 BFS 一样同样需要对已经遍历过的节点进行标记。

1. 查找最大的连通面积

695. Max Area of Island (Medium)

Leetcodeopen in new window / 力扣open in new window

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]

private int m, n;
private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

public int maxAreaOfIsland(int[][] grid) {
    if (grid == null || grid.length == 0) {
        return 0;
    }
    m = grid.length;
    n = grid[0].length;
    int maxArea = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            maxArea = Math.max(maxArea, dfs(grid, i, j));
        }
    }
    return maxArea;
}

private int dfs(int[][] grid, int r, int c) {
    if (r < 0 || r >= m || c < 0 || c >= n || grid[r][c] == 0) {
        return 0;
    }
    grid[r][c] = 0;
    int area = 1;
    for (int[] d : direction) {
        area += dfs(grid, r + d[0], c + d[1]);
    }
    return area;
}

2. 矩阵中的连通分量数目

200. Number of Islands (Medium)

Leetcodeopen in new window / 力扣open in new window

Input:
11000
11000
00100
00011

Output: 3

可以将矩阵表示看成一张有向图。

private int m, n;
private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

public int numIslands(char[][] grid) {
    if (grid == null || grid.length == 0) {
        return 0;
    }
    m = grid.length;
    n = grid[0].length;
    int islandsNum = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid[i][j] != '0') {
                dfs(grid, i, j);
                islandsNum++;
            }
        }
    }
    return islandsNum;
}

private void dfs(char[][] grid, int i, int j) {
    if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0') {
        return;
    }
    grid[i][j] = '0';
    for (int[] d : direction) {
        dfs(grid, i + d[0], j + d[1]);
    }
}

3. 好友关系的连通分量数目

547. Friend Circles (Medium)

Leetcodeopen in new window / 力扣open in new window

Input:
[[1,1,0],
 [1,1,0],
 [0,0,1]]

Output: 2

Explanation:The 0th and 1st students are direct friends, so they are in a friend circle.
The 2nd student himself is in a friend circle. So return 2.

题目描述：好友关系可以看成是一个无向图，例如第 0 个人与第 1 个人是好友，那么 M[0][1] 和 M[1][0] 的值都为 1。

private int n;

public int findCircleNum(int[][] M) {
    n = M.length;
    int circleNum = 0;
    boolean[] hasVisited = new boolean[n];
    for (int i = 0; i < n; i++) {
        if (!hasVisited[i]) {
            dfs(M, i, hasVisited);
            circleNum++;
        }
    }
    return circleNum;
}

private void dfs(int[][] M, int i, boolean[] hasVisited) {
    hasVisited[i] = true;
    for (int k = 0; k < n; k++) {
        if (M[i][k] == 1 && !hasVisited[k]) {
            dfs(M, k, hasVisited);
        }
    }
}

4. 填充封闭区域

130. Surrounded Regions (Medium)

Leetcodeopen in new window / 力扣open in new window

For example,
X X X X
X O O X
X X O X
X O X X

After running your function, the board should be:
X X X X
X X X X
X X X X
X O X X

题目描述：使被 'X' 包围的 'O' 转换为 'X'。

先填充最外侧，剩下的就是里侧了。

private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
private int m, n;

public void solve(char[][] board) {
    if (board == null || board.length == 0) {
        return;
    }

    m = board.length;
    n = board[0].length;

    for (int i = 0; i < m; i++) {
        dfs(board, i, 0);
        dfs(board, i, n - 1);
    }
    for (int i = 0; i < n; i++) {
        dfs(board, 0, i);
        dfs(board, m - 1, i);
    }

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (board[i][j] == 'T') {
                board[i][j] = 'O';
            } else if (board[i][j] == 'O') {
                board[i][j] = 'X';
            }
        }
    }
}

private void dfs(char[][] board, int r, int c) {
    if (r < 0 || r >= m || c < 0 || c >= n || board[r][c] != 'O') {
        return;
    }
    board[r][c] = 'T';
    for (int[] d : direction) {
        dfs(board, r + d[0], c + d[1]);
    }
}

5. 能到达的太平洋和大西洋的区域

417. Pacific Atlantic Water Flow (Medium)

Leetcodeopen in new window / 力扣open in new window

Given the following 5x5 matrix:

  Pacific ~   ~   ~   ~   ~
       ~  1   2   2   3  (5) *
       ~  3   2   3  (4) (4) *
       ~  2   4  (5)  3   1  *
       ~ (6) (7)  1   4   5  *
       ~ (5)  1   1   2   4  *
          *   *   *   *   * Atlantic

Return:
[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).

左边和上边是太平洋，右边和下边是大西洋，内部的数字代表海拔，海拔高的地方的水能够流到低的地方，求解水能够流到太平洋和大西洋的所有位置。

private int m, n;
private int[][] matrix;
private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

public List<List<Integer>> pacificAtlantic(int[][] matrix) {
    List<List<Integer>> ret = new ArrayList<>();
    if (matrix == null || matrix.length == 0) {
        return ret;
    }

    m = matrix.length;
    n = matrix[0].length;
    this.matrix = matrix;
    boolean[][] canReachP = new boolean[m][n];
    boolean[][] canReachA = new boolean[m][n];

    for (int i = 0; i < m; i++) {
        dfs(i, 0, canReachP);
        dfs(i, n - 1, canReachA);
    }
    for (int i = 0; i < n; i++) {
        dfs(0, i, canReachP);
        dfs(m - 1, i, canReachA);
    }

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (canReachP[i][j] && canReachA[i][j]) {
                ret.add(Arrays.asList(i, j));
            }
        }
    }

    return ret;
}

private void dfs(int r, int c, boolean[][] canReach) {
    if (canReach[r][c]) {
        return;
    }
    canReach[r][c] = true;
    for (int[] d : direction) {
        int nextR = d[0] + r;
        int nextC = d[1] + c;
        if (nextR < 0 || nextR >= m || nextC < 0 || nextC >= n
                || matrix[r][c] > matrix[nextR][nextC]) {

            continue;
        }
        dfs(nextR, nextC, canReach);
    }
}

Backtracking

Backtracking（回溯）属于 DFS。

普通 DFS 主要用在 可达性问题 ，这种问题只需要执行到特点的位置然后返回即可。
而 Backtracking 主要用于求解 排列组合 问题，例如有 { 'a','b','c' } 三个字符，求解所有由这三个字符排列得到的字符串，这种问题在执行到特定的位置返回之后还会继续执行求解过程。

因为 Backtracking 不是立即返回，而要继续求解，因此在程序实现时，需要注意对元素的标记问题：

在访问一个新元素进入新的递归调用时，需要将新元素标记为已经访问，这样才能在继续递归调用时不用重复访问该元素；
但是在递归返回时，需要将元素标记为未访问，因为只需要保证在一个递归链中不同时访问一个元素，可以访问已经访问过但是不在当前递归链中的元素。

1. 数字键盘组合

17. Letter Combinations of a Phone Number (Medium)

Leetcodeopen in new window / 力扣open in new window

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

private static final String[] KEYS = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

public List<String> letterCombinations(String digits) {
    List<String> combinations = new ArrayList<>();
    if (digits == null || digits.length() == 0) {
        return combinations;
    }
    doCombination(new StringBuilder(), combinations, digits);
    return combinations;
}

private void doCombination(StringBuilder prefix, List<String> combinations, final String digits) {
    if (prefix.length() == digits.length()) {
        combinations.add(prefix.toString());
        return;
    }
    int curDigits = digits.charAt(prefix.length()) - '0';
    String letters = KEYS[curDigits];
    for (char c : letters.toCharArray()) {
        prefix.append(c);                         // 添加
        doCombination(prefix, combinations, digits);
        prefix.deleteCharAt(prefix.length() - 1); // 删除
    }
}

2. IP 地址划分

93. Restore IP Addresses(Medium)

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Given "25525511135",
return ["255.255.11.135", "255.255.111.35"].

public List<String> restoreIpAddresses(String s) {
    List<String> addresses = new ArrayList<>();
    StringBuilder tempAddress = new StringBuilder();
    doRestore(0, tempAddress, addresses, s);
    return addresses;
}

private void doRestore(int k, StringBuilder tempAddress, List<String> addresses, String s) {
    if (k == 4 || s.length() == 0) {
        if (k == 4 && s.length() == 0) {
            addresses.add(tempAddress.toString());
        }
        return;
    }
    for (int i = 0; i < s.length() && i <= 2; i++) {
        if (i != 0 && s.charAt(0) == '0') {
            break;
        }
        String part = s.substring(0, i + 1);
        if (Integer.valueOf(part) <= 255) {
            if (tempAddress.length() != 0) {
                part = "." + part;
            }
            tempAddress.append(part);
            doRestore(k + 1, tempAddress, addresses, s.substring(i + 1));
            tempAddress.delete(tempAddress.length() - part.length(), tempAddress.length());
        }
    }
}

3. 在矩阵中寻找字符串

79. Word Search (Medium)

Leetcodeopen in new window / 力扣open in new window

For example,
Given board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

private final static int[][] direction = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
private int m;
private int n;

public boolean exist(char[][] board, String word) {
    if (word == null || word.length() == 0) {
        return true;
    }
    if (board == null || board.length == 0 || board[0].length == 0) {
        return false;
    }

    m = board.length;
    n = board[0].length;
    boolean[][] hasVisited = new boolean[m][n];

    for (int r = 0; r < m; r++) {
        for (int c = 0; c < n; c++) {
            if (backtracking(0, r, c, hasVisited, board, word)) {
                return true;
            }
        }
    }

    return false;
}

private boolean backtracking(int curLen, int r, int c, boolean[][] visited, final char[][] board, final String word) {
    if (curLen == word.length()) {
        return true;
    }
    if (r < 0 || r >= m || c < 0 || c >= n
            || board[r][c] != word.charAt(curLen) || visited[r][c]) {

        return false;
    }

    visited[r][c] = true;

    for (int[] d : direction) {
        if (backtracking(curLen + 1, r + d[0], c + d[1], visited, board, word)) {
            return true;
        }
    }

    visited[r][c] = false;

    return false;
}

4. 输出二叉树中所有从根到叶子的路径

257. Binary Tree Paths (Easy)

Leetcodeopen in new window / 力扣open in new window

["1->2->5", "1->3"]


public List<String> binaryTreePaths(TreeNode root) {
    List<String> paths = new ArrayList<>();
    if (root == null) {
        return paths;
    }
    List<Integer> values = new ArrayList<>();
    backtracking(root, values, paths);
    return paths;
}

private void backtracking(TreeNode node, List<Integer> values, List<String> paths) {
    if (node == null) {
        return;
    }
    values.add(node.val);
    if (isLeaf(node)) {
        paths.add(buildPath(values));
    } else {
        backtracking(node.left, values, paths);
        backtracking(node.right, values, paths);
    }
    values.remove(values.size() - 1);
}

private boolean isLeaf(TreeNode node) {
    return node.left == null && node.right == null;
}

private String buildPath(List<Integer> values) {
    StringBuilder str = new StringBuilder();
    for (int i = 0; i < values.size(); i++) {
        str.append(values.get(i));
        if (i != values.size() - 1) {
            str.append("->");
        }
    }
    return str.toString();
}

5. 排列

46. Permutations (Medium)

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[1,2,3] have the following permutations:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

public List<List<Integer>> permute(int[] nums) {
    List<List<Integer>> permutes = new ArrayList<>();
    List<Integer> permuteList = new ArrayList<>();
    boolean[] hasVisited = new boolean[nums.length];
    backtracking(permuteList, permutes, hasVisited, nums);
    return permutes;
}

private void backtracking(List<Integer> permuteList, List<List<Integer>> permutes, boolean[] visited, final int[] nums) {
    if (permuteList.size() == nums.length) {
        permutes.add(new ArrayList<>(permuteList)); // 重新构造一个 List
        return;
    }
    for (int i = 0; i < visited.length; i++) {
        if (visited[i]) {
            continue;
        }
        visited[i] = true;
        permuteList.add(nums[i]);
        backtracking(permuteList, permutes, visited, nums);
        permuteList.remove(permuteList.size() - 1);
        visited[i] = false;
    }
}

6. 含有相同元素求排列

47. Permutations II (Medium)

Leetcodeopen in new window / 力扣open in new window

[1,1,2] have the following unique permutations:
[[1,1,2], [1,2,1], [2,1,1]]

数组元素可能含有相同的元素，进行排列时就有可能出现重复的排列，要求重复的排列只返回一个。

在实现上，和 Permutations 不同的是要先排序，然后在添加一个元素时，判断这个元素是否等于前一个元素，如果等于，并且前一个元素还未访问，那么就跳过这个元素。

public List<List<Integer>> permuteUnique(int[] nums) {
    List<List<Integer>> permutes = new ArrayList<>();
    List<Integer> permuteList = new ArrayList<>();
    Arrays.sort(nums);  // 排序
    boolean[] hasVisited = new boolean[nums.length];
    backtracking(permuteList, permutes, hasVisited, nums);
    return permutes;
}

private void backtracking(List<Integer> permuteList, List<List<Integer>> permutes, boolean[] visited, final int[] nums) {
    if (permuteList.size() == nums.length) {
        permutes.add(new ArrayList<>(permuteList));
        return;
    }

    for (int i = 0; i < visited.length; i++) {
        if (i != 0 && nums[i] == nums[i - 1] && !visited[i - 1]) {
            continue;  // 防止重复
        }
        if (visited[i]){
            continue;
        }
        visited[i] = true;
        permuteList.add(nums[i]);
        backtracking(permuteList, permutes, visited, nums);
        permuteList.remove(permuteList.size() - 1);
        visited[i] = false;
    }
}

7. 组合

77. Combinations (Medium)

Leetcodeopen in new window / 力扣open in new window

If n = 4 and k = 2, a solution is:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

public List<List<Integer>> combine(int n, int k) {
    List<List<Integer>> combinations = new ArrayList<>();
    List<Integer> combineList = new ArrayList<>();
    backtracking(combineList, combinations, 1, k, n);
    return combinations;
}

private void backtracking(List<Integer> combineList, List<List<Integer>> combinations, int start, int k, final int n) {
    if (k == 0) {
        combinations.add(new ArrayList<>(combineList));
        return;
    }
    for (int i = start; i <= n - k + 1; i++) {  // 剪枝
        combineList.add(i);
        backtracking(combineList, combinations, i + 1, k - 1, n);
        combineList.remove(combineList.size() - 1);
    }
}

8. 组合求和

39. Combination Sum (Medium)

Leetcodeopen in new window / 力扣open in new window

given candidate set [2, 3, 6, 7] and target 7,
A solution set is:
[[7],[2, 2, 3]]

public List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<List<Integer>> combinations = new ArrayList<>();
    backtracking(new ArrayList<>(), combinations, 0, target, candidates);
    return combinations;
}

private void backtracking(List<Integer> tempCombination, List<List<Integer>> combinations,
                          int start, int target, final int[] candidates) {

    if (target == 0) {
        combinations.add(new ArrayList<>(tempCombination));
        return;
    }
    for (int i = start; i < candidates.length; i++) {
        if (candidates[i] <= target) {
            tempCombination.add(candidates[i]);
            backtracking(tempCombination, combinations, i, target - candidates[i], candidates);
            tempCombination.remove(tempCombination.size() - 1);
        }
    }
}

9. 含有相同元素的组合求和

40. Combination Sum II (Medium)

Leetcodeopen in new window / 力扣open in new window

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    List<List<Integer>> combinations = new ArrayList<>();
    Arrays.sort(candidates);
    backtracking(new ArrayList<>(), combinations, new boolean[candidates.length], 0, target, candidates);
    return combinations;
}

private void backtracking(List<Integer> tempCombination, List<List<Integer>> combinations,
                          boolean[] hasVisited, int start, int target, final int[] candidates) {

    if (target == 0) {
        combinations.add(new ArrayList<>(tempCombination));
        return;
    }
    for (int i = start; i < candidates.length; i++) {
        if (i != 0 && candidates[i] == candidates[i - 1] && !hasVisited[i - 1]) {
            continue;
        }
        if (candidates[i] <= target) {
            tempCombination.add(candidates[i]);
            hasVisited[i] = true;
            backtracking(tempCombination, combinations, hasVisited, i + 1, target - candidates[i], candidates);
            hasVisited[i] = false;
            tempCombination.remove(tempCombination.size() - 1);
        }
    }
}

10. 1-9 数字的组合求和

216. Combination Sum III (Medium)

Leetcodeopen in new window / 力扣open in new window

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

从 1-9 数字中选出 k 个数不重复的数，使得它们的和为 n。

public List<List<Integer>> combinationSum3(int k, int n) {
    List<List<Integer>> combinations = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    backtracking(k, n, 1, path, combinations);
    return combinations;
}

private void backtracking(int k, int n, int start,
                          List<Integer> tempCombination, List<List<Integer>> combinations) {

    if (k == 0 && n == 0) {
        combinations.add(new ArrayList<>(tempCombination));
        return;
    }
    if (k == 0 || n == 0) {
        return;
    }
    for (int i = start; i <= 9; i++) {
        tempCombination.add(i);
        backtracking(k - 1, n - i, i + 1, tempCombination, combinations);
        tempCombination.remove(tempCombination.size() - 1);
    }
}

11. 子集

78. Subsets (Medium)

Leetcodeopen in new window / 力扣open in new window

找出集合的所有子集，子集不能重复，[1, 2] 和 [2, 1] 这种子集算重复

public List<List<Integer>> subsets(int[] nums) {
    List<List<Integer>> subsets = new ArrayList<>();
    List<Integer> tempSubset = new ArrayList<>();
    for (int size = 0; size <= nums.length; size++) {
        backtracking(0, tempSubset, subsets, size, nums); // 不同的子集大小
    }
    return subsets;
}

private void backtracking(int start, List<Integer> tempSubset, List<List<Integer>> subsets,
                          final int size, final int[] nums) {

    if (tempSubset.size() == size) {
        subsets.add(new ArrayList<>(tempSubset));
        return;
    }
    for (int i = start; i < nums.length; i++) {
        tempSubset.add(nums[i]);
        backtracking(i + 1, tempSubset, subsets, size, nums);
        tempSubset.remove(tempSubset.size() - 1);
    }
}

12. 含有相同元素求子集

90. Subsets II (Medium)

Leetcodeopen in new window / 力扣open in new window

For example,
If nums = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

public List<List<Integer>> subsetsWithDup(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> subsets = new ArrayList<>();
    List<Integer> tempSubset = new ArrayList<>();
    boolean[] hasVisited = new boolean[nums.length];
    for (int size = 0; size <= nums.length; size++) {
        backtracking(0, tempSubset, subsets, hasVisited, size, nums); // 不同的子集大小
    }
    return subsets;
}

private void backtracking(int start, List<Integer> tempSubset, List<List<Integer>> subsets, boolean[] hasVisited,
                          final int size, final int[] nums) {

    if (tempSubset.size() == size) {
        subsets.add(new ArrayList<>(tempSubset));
        return;
    }
    for (int i = start; i < nums.length; i++) {
        if (i != 0 && nums[i] == nums[i - 1] && !hasVisited[i - 1]) {
            continue;
        }
        tempSubset.add(nums[i]);
        hasVisited[i] = true;
        backtracking(i + 1, tempSubset, subsets, hasVisited, size, nums);
        hasVisited[i] = false;
        tempSubset.remove(tempSubset.size() - 1);
    }
}

13. 分割字符串使得每个部分都是回文数

131. Palindrome Partitioning (Medium)

Leetcodeopen in new window / 力扣open in new window

For example, given s = "aab",
Return

[
  ["aa","b"],
  ["a","a","b"]
]

public List<List<String>> partition(String s) {
    List<List<String>> partitions = new ArrayList<>();
    List<String> tempPartition = new ArrayList<>();
    doPartition(s, partitions, tempPartition);
    return partitions;
}

private void doPartition(String s, List<List<String>> partitions, List<String> tempPartition) {
    if (s.length() == 0) {
        partitions.add(new ArrayList<>(tempPartition));
        return;
    }
    for (int i = 0; i < s.length(); i++) {
        if (isPalindrome(s, 0, i)) {
            tempPartition.add(s.substring(0, i + 1));
            doPartition(s.substring(i + 1), partitions, tempPartition);
            tempPartition.remove(tempPartition.size() - 1);
        }
    }
}

private boolean isPalindrome(String s, int begin, int end) {
    while (begin < end) {
        if (s.charAt(begin++) != s.charAt(end--)) {
            return false;
        }
    }
    return true;
}

14. 数独

37. Sudoku Solver (Hard)

Leetcodeopen in new window / 力扣open in new window

private boolean[][] rowsUsed = new boolean[9][10];
private boolean[][] colsUsed = new boolean[9][10];
private boolean[][] cubesUsed = new boolean[9][10];
private char[][] board;

public void solveSudoku(char[][] board) {
    this.board = board;
    for (int i = 0; i < 9; i++)
        for (int j = 0; j < 9; j++) {
            if (board[i][j] == '.') {
                continue;
            }
            int num = board[i][j] - '0';
            rowsUsed[i][num] = true;
            colsUsed[j][num] = true;
            cubesUsed[cubeNum(i, j)][num] = true;
        }
        backtracking(0, 0);
}

private boolean backtracking(int row, int col) {
    while (row < 9 && board[row][col] != '.') {
        row = col == 8 ? row + 1 : row;
        col = col == 8 ? 0 : col + 1;
    }
    if (row == 9) {
        return true;
    }
    for (int num = 1; num <= 9; num++) {
        if (rowsUsed[row][num] || colsUsed[col][num] || cubesUsed[cubeNum(row, col)][num]) {
            continue;
        }
        rowsUsed[row][num] = colsUsed[col][num] = cubesUsed[cubeNum(row, col)][num] = true;
        board[row][col] = (char) (num + '0');
        if (backtracking(row, col)) {
            return true;
        }
        board[row][col] = '.';
        rowsUsed[row][num] = colsUsed[col][num] = cubesUsed[cubeNum(row, col)][num] = false;
    }
    return false;
}

private int cubeNum(int i, int j) {
    int r = i / 3;
    int c = j / 3;
    return r * 3 + c;
}

15. N 皇后

51. N-Queens (Hard)

Leetcodeopen in new window / 力扣open in new window

在 n*n 的矩阵中摆放 n 个皇后，并且每个皇后不能在同一行，同一列，同一对角线上，求所有的 n 皇后的解。

一行一行地摆放，在确定一行中的那个皇后应该摆在哪一列时，需要用三个标记数组来确定某一列是否合法，这三个标记数组分别为：列标记数组、45 度对角线标记数组和 135 度对角线标记数组。

45 度对角线标记数组的长度为 2 * n - 1，通过下图可以明确 (r, c) 的位置所在的数组下标为 r + c。

![img](https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/9c422923-1447-4a3b-a4e1-97e663738187.jpg" width="300px">

135 度对角线标记数组的长度也是 2 * n - 1，(r, c) 的位置所在的数组下标为 n - 1 - (r - c)。

![img](https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/7a85e285-e152-4116-b6dc-3fab27ba9437.jpg" width="300px">

private List<List<String>> solutions;
private char[][] nQueens;
private boolean[] colUsed;
private boolean[] diagonals45Used;
private boolean[] diagonals135Used;
private int n;

public List<List<String>> solveNQueens(int n) {
    solutions = new ArrayList<>();
    nQueens = new char[n][n];
    for (int i = 0; i < n; i++) {
        Arrays.fill(nQueens[i], '.');
    }
    colUsed = new boolean[n];
    diagonals45Used = new boolean[2 * n - 1];
    diagonals135Used = new boolean[2 * n - 1];
    this.n = n;
    backtracking(0);
    return solutions;
}

private void backtracking(int row) {
    if (row == n) {
        List<String> list = new ArrayList<>();
        for (char[] chars : nQueens) {
            list.add(new String(chars));
        }
        solutions.add(list);
        return;
    }

    for (int col = 0; col < n; col++) {
        int diagonals45Idx = row + col;
        int diagonals135Idx = n - 1 - (row - col);
        if (colUsed[col] || diagonals45Used[diagonals45Idx] || diagonals135Used[diagonals135Idx]) {
            continue;
        }
        nQueens[row][col] = 'Q';
        colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = true;
        backtracking(row + 1);
        colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = false;
        nQueens[row][col] = '.';
    }
}

Leetcode 题解 - 搜索

# Leetcode 题解 - 搜索

# BFS

# 1. 计算在网格中从原点到特定点的最短路径长度

# 2. 组成整数的最小平方数数量

# 3. 最短单词路径

# DFS

# 1. 查找最大的连通面积

# 2. 矩阵中的连通分量数目

# 3. 好友关系的连通分量数目

# 4. 填充封闭区域

# 5. 能到达的太平洋和大西洋的区域

# Backtracking

# 1. 数字键盘组合

# 2. IP 地址划分

# 3. 在矩阵中寻找字符串

# 4. 输出二叉树中所有从根到叶子的路径

# 5. 排列

# 6. 含有相同元素求排列

# 7. 组合

# 8. 组合求和

# 9. 含有相同元素的组合求和

# 10. 1-9 数字的组合求和

# 11. 子集

# 12. 含有相同元素求子集

# 13. 分割字符串使得每个部分都是回文数

# 14. 数独

# 15. N 皇后

Leetcode 题解 - 搜索

BFS

1. 计算在网格中从原点到特定点的最短路径长度

2. 组成整数的最小平方数数量

3. 最短单词路径

DFS

1. 查找最大的连通面积

2. 矩阵中的连通分量数目

3. 好友关系的连通分量数目

4. 填充封闭区域

5. 能到达的太平洋和大西洋的区域

Backtracking

1. 数字键盘组合

2. IP 地址划分

3. 在矩阵中寻找字符串

4. 输出二叉树中所有从根到叶子的路径

5. 排列

6. 含有相同元素求排列

7. 组合

8. 组合求和

9. 含有相同元素的组合求和

10. 1-9 数字的组合求和

11. 子集

12. 含有相同元素求子集

13. 分割字符串使得每个部分都是回文数

14. 数独

15. N 皇后