CS-Notes/notes/Leetcode 题解 - 搜索.md

• BFS
  • 计算在网格中从原点到特定点的最短路径长度
  • 组成整数的最小平方数数量
  • 最短单词路径
• DFS
  • 查找最大的连通面积
  • 矩阵中的连通分量数目
  • 好友关系的连通分量数目
  • 填充封闭区域
  • 能到达的太平洋和大西洋的区域
• Backtracking
  • 数字键盘组合
  • IP 地址划分
  • 在矩阵中寻找字符串
  • 输出二叉树中所有从根到叶子的路径
  • 排列
  • 含有相同元素求排列
  • 组合
  • 组合求和
  • 含有相同元素的组合求和
  • 1-9 数字的组合求和
  • 子集
  • 含有相同元素求子集
  • 分割字符串使得每个部分都是回文数
  • 数独
  • N 皇后

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。

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。

计算在网格中从原点到特定点的最短路径长度

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

1 表示可以经过某个位置，求解从 (0, 0) 位置到 (tr, tc) 位置的最短路径长度。

public int minPathLength(int[][] grids, int tr, int tc) {
    final int[][] direction = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    final 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();
            grids[cr][cc] = 0; // 标记
            for (int[] d : direction) {
                int nr = cr + d[0], nc = cc + d[1];
                if (nr < 0 || nr >= m || nc < 0 || nc >= n || grids[nr][nc] == 0) {
                    continue;
                }
                if (nr == tr && nc == tc) {
                    return pathLength;
                }
                queue.add(new Pair<>(nr, nc));
            }
        }
    }
    return -1;
}

组成整数的最小平方数数量

279. Perfect Squares (Medium)

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;
}

最短单词路径

127. Word Ladder (Medium)

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。

查找最大的连通面积

695. Max Area of Island (Medium)

[[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;
}

矩阵中的连通分量数目

200. Number of Islands (Medium)

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]);
    }
}

好友关系的连通分量数目

547. Friend Circles (Medium)

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);
        }
    }
}

填充封闭区域

130. Surrounded Regions (Medium)

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]);
    }
}

能到达的太平洋和大西洋的区域

417. Pacific Atlantic Water Flow (Medium)

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<int[]> pacificAtlantic(int[][] matrix) {
    List<int[]> 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(new int[]{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 不是立即就返回，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。

数字键盘组合

17. Letter Combinations of a Phone Number (Medium)

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
    }
}

IP 地址划分

93. Restore IP Addresses(Medium)

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());
        }
    }
}

在矩阵中寻找字符串

79. Word Search (Medium)

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;
}

输出二叉树中所有从根到叶子的路径

257. Binary Tree Paths (Easy)

  1
 /  \
2    3
 \
  5

["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();
}

排列

46. Permutations (Medium)

[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;
    }
}

含有相同元素求排列

47. Permutations II (Medium)

[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;
    }
}

组合

77. Combinations (Medium)

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);
    }
}

组合求和

39. Combination Sum (Medium)

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);
        }
    }
}

含有相同元素的组合求和

40. Combination Sum II (Medium)

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);
        }
    }
}

1-9 数字的组合求和

216. Combination Sum III (Medium)

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);
    }
}

子集

78. Subsets (Medium)

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);
    }
}

含有相同元素求子集

90. Subsets II (Medium)

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);
    }
}

分割字符串使得每个部分都是回文数

131. Palindrome Partitioning (Medium)

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;
}

数独

37. Sudoku Solver (Hard)

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;
}

N 皇后

51. N-Queens (Hard)

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。

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)。

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] = '.';
    }
}

🎨️更多精彩内容请关注公众号 CyC2018，在公众号后台回复关键字 资料 可领取技术面试复习大纲，这份大纲是我花了一整年时间整理的面试知识点列表，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点。我基本是按照这份大纲来进行复习的，这份大纲对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。