LeetCode——Backtracking

末蓝、 2023-06-19 09:38 31阅读 0赞

# Backtracking #

--------------------

## 目录 ##

1.  Backtracking
2.  数字键盘组合
3.  IP 地址划分
4.  在矩阵中寻找字符串
5.  输出二叉树中所有从根到叶子的路径
6.  排列
7.  含有相同元素求排列
8.  组合
9.  组合求和
10. 含有相同元素的组合求和
11. 1-9 数字的组合求和
12. 子集
13. 含有相同元素求子集
14. 分割字符串使得每个部分都是回文数
15. 数独
16. N皇后

--------------------

### 1. 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。

--------------------

### 2. 数字键盘组合 ###

[https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number][https_leetcode-cn.com_problems_letter-combinations-of-a-phone-number]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70]

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 num = digits.charAt(prefix.length()) - '0';
            String key = KEYS[num];
            for (char c : key.toCharArray()) {
        
                prefix.append(c);
                doCombination(prefix,combinations,digits);
                prefix.deleteCharAt(prefix.length()-1);
            }
        }

--------------------

### 3. IP 地址划分 ###

[https://leetcode-cn.com/problems/restore-ip-addresses][https_leetcode-cn.com_problems_restore-ip-addresses]

![在这里插入图片描述][20191205171051876.png]

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) {
        
            //整体分成4个段
            if (k == 4 || s.length() == 0) {
        
                if (k == 4 && s.length() == 0) {
        
                    addresses.add(tempAddress.toString());
                }
                return;
            }
            //每个段的判断
            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());
                }
            }
        }

--------------------

### 4. 在矩阵中寻找字符串 ###

[https://leetcode-cn.com/problems/word-search][https_leetcode-cn.com_problems_word-search]

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 1]  
题目描述：给定一个二维网格和一个单词，找出该单词是否存在于网格中。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

private final static int[][] direction = {
        {
        1, 0}, {
        -1, 0}, {
        0, 1}, {
        0, -1}};
        private int m, 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[][] visited = new boolean[m][n];
            for (int r = 0; r < m; r++) {
        
                for (int c = 0; c < n; c++) {
        
                    if (backtracking(0, r, c, visited, 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;
        }

--------------------

### 5. 输出二叉树中所有从根到叶子的路径 ###

[https://leetcode.com/problems/binary-tree-paths][https_leetcode.com_problems_binary-tree-paths]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 2]

public static class TreeNode {
        
            int val = 0;
            TreeNode left = null;
            TreeNode right = null;
    
            public TreeNode(int val) {
        
                this.val = val;
            }
        }
    
        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 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();
        }
    
        private boolean isLeaf(TreeNode node) {
        
            return node.left == null && node.right == null;
        }

--------------------

### 6. 排列 ###

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 3]  
题目描述：给定一个没有重复数字的序列，返回其所有可能的全排列。

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, 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;
            }
        }

--------------------

### 7. 含有相同元素求排列 ###

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 4]  
题目描述：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);
            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, 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;
            }
        }

--------------------

### 8. 组合 ###

[https://leetcode-cn.com/problems/combinations][https_leetcode-cn.com_problems_combinations]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 5]

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

--------------------

### 9. 组合求和 ###

[https://leetcode.com/problems/combination-sum/description/][https_leetcode.com_problems_combination-sum_description]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 6]  
题目描述：给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。

说明：

*  所有数字（包括 target）都是正整数。
 *  解集不能包含重复的组合。

public List<List<Integer>> combiationSum(int[] candidates, int target) {
        
            List<List<Integer>> combinations = new ArrayList<>();
            backtracking(new ArrayList<>(), combinations, 0, target, candidates);
            return combinations;
    
        }
    
        private void backtracking(ArrayList<Integer> tempCombination, List<List<Integer>> combinations, int start, int target, 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);
                }
            }
        }

--------------------

### 10. 含有相同元素的组合求和 ###

[https://leetcode.com/problems/combination-sum-ii/description/][https_leetcode.com_problems_combination-sum-ii_description]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 7]  
题目描述：给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用一次。

说明：

*  所有数字（包括目标数）都是正整数。
 *  解集不能包含重复的组合。

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        
            List<List<Integer>> combinations = new ArrayList<>();
            Arrays.sort(candidates);
            backtracking(combinations, new ArrayList<>(), new boolean[candidates.length], 0, target, candidates);
            return combinations;
        }
    
        private void backtracking(List<List<Integer>> combinations, List<Integer> combinationList, boolean[] visited, int start, int target, int[] candidates) {
        
            if (target == 0) {
        
                combinations.add(new ArrayList<>(combinationList));
                return;
            }
    
            for (int i = start; i < candidates.length; i++) {
        
                if (i != 0 && candidates[i] == candidates[i - 1] && !visited[i - 1]) {
        
                    continue;
                }
                if (candidates[i] <= target) {
        
                    combinationList.add(candidates[i]);
                    visited[i] = true;
                    backtracking(combinations, combinationList, visited, i + 1, target - candidates[i], candidates);
                    visited[i] = false;
                    combinationList.remove(combinationList.size() - 1);
                }
            }
        }

--------------------

### 11. 1-9 数字的组合求和 ###

[https://leetcode.com/problems/combination-sum-iii/description/][https_leetcode.com_problems_combination-sum-iii_description]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 8]  
题目描述：找出所有相加之和为 n 的 k 个数的组合。组合中只允许含有 1 - 9 的正整数，并且每种组合中不存在重复的数字。

说明：

*  所有数字都是正整数。
 *  解集不能包含重复的组合。

public List<List<Integer>> combinationSum3(int k, int n) {
        
            List<List<Integer>> combinations = new ArrayList<>();
            List<Integer> combinationList = new ArrayList<>();
            backtracking(combinationList, combinations, 1, k, n);
            return combinations;
        }
    
        private void backtracking(List<Integer> combinationList, List<List<Integer>> combinations, int start, int k, int target) {
        
            if (k == 0 && target == 0) {
        
                combinations.add(new ArrayList<>(combinationList));
                return;
            }
            if (k == 0 || target == 0) {
        
                return;
            }
            for (int i = start; i <= 9; i++) {
        
                combinationList.add(i);
                backtracking(combinationList, combinations, i + 1, k - 1, target - i);
                combinationList.remove(combinationList.size() - 1);
            }
        }

--------------------

### 12. 子集 ###

[https://leetcode.com/problems/subsets-ii/description/][https_leetcode.com_problems_subsets-ii_description]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 9]  
题目描述：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(subsets, tempSubset, size, 0, nums);
            }
            return subsets;
        }
    
        private void backtracking(List<List<Integer>> subsets, List<Integer> tempSubset, int size, int start, int[] nums) {
        
            if (size == 0) {
        
                subsets.add(new ArrayList<>(tempSubset));
            }
    
            for (int i = start; i < nums.length; i++) {
        
                tempSubset.add(nums[i]);
                backtracking(subsets, tempSubset, size - 1, i + 1, nums);
                tempSubset.remove(tempSubset.size() - 1);
            }
        }

--------------------

### 13. 含有相同元素求子集 ###

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 10]  
题目描述：给定一个可能包含重复数的整数集合，返回所有可能的子集(幂集)。

注意:解决方案集不能包含重复的子集。

--------------------

### 14. 分割字符串使得每个部分都是回文数 ###

[https://leetcode-cn.com/problems/palindrome-partitioning][https_leetcode-cn.com_problems_palindrome-partitioning]  
![在这里插入图片描述][20191208122902703.png]  
题目描述：给定一个字符串 s，将 s 分割成一些子串，使每个子串都是回文串。

返回 s 所有可能的分割方案。

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;
        }

--------------------

### 15. 数独 ###

[https://leetcode-cn.com/problems/sudoku-solver/solution/jie-shu-du-by-leetcode/][https_leetcode-cn.com_problems_sudoku-solver_solution_jie-shu-du-by-leetcode]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 11]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 12]  
题目描述：编写一个程序，通过已填充的空格来解决数独问题。

一个数独的解法需遵循如下规则：

*  数字 1-9 在每一行只能出现一次。
 *  数字 1-9 在每一列只能出现一次。
 *  数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
 *  空白格用 ‘.’ 表示。

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[cubes(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[cubes(row, col)][num]) {
        
                    continue;
                }
                rowsUsed[row][num] = colsUsed[col][num] = cubesUsed[cubes(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[cubes(row, col)][num] == false;
            }
            return false;
        }
    
        private int cubes(int i, int j) {
        
            int r = i / 3;
            int c = j / 3;
            return r * 3 + c;
        }

--------------------

### 16. N皇后 ###

[https://leetcode-cn.com/problems/n-queens/solution/nhuang-hou-by-leetcode/][https_leetcode-cn.com_problems_n-queens_solution_nhuang-hou-by-leetcode]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 13]  
题目描述：n 皇后问题研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击。

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 14]

上图为 8 皇后问题的一种解法。

给定一个整数 n，返回所有不同的 n 皇后问题的解决方案。

每一种解法包含一个明确的 n 皇后问题的棋子放置方案，该方案中 ‘Q’ 和 ‘.’ 分别代表了皇后和空位。

private List<List<String>> solutions;
        private char[][] nQueues;
        private boolean[] colUsed;
        private boolean[] diagonals45Used;
        private boolean[] diagonals135Used;
        private int n;
    
        public List<List<String>> solveNQueues(int n) {
        
            solutions = new ArrayList<>();
            nQueues = new char[n][n];
            for (int i = 0; i < n; i++) {
        
                Arrays.fill(nQueues[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 : nQueues) {
        
                    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;
                }
                nQueues[row][col] = 'Q';
                colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = true;
                backtracking(row + 1);
                colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = false;
                nQueues[row][col] = '.';
            }
        }

[https_leetcode-cn.com_problems_letter-combinations-of-a-phone-number]: https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/description/?utm_source=LCUS&utm_medium=ip_redirect_q_uns&utm_campaign=transfer2china
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70]: https://img-blog.csdnimg.cn/20191205151658204.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode-cn.com_problems_restore-ip-addresses]: https://leetcode-cn.com/problems/restore-ip-addresses/description/?utm_source=LCUS&utm_medium=ip_redirect_q_uns&utm_campaign=transfer2china
[20191205171051876.png]: https://img-blog.csdnimg.cn/20191205171051876.png
[https_leetcode-cn.com_problems_word-search]: https://leetcode-cn.com/problems/word-search/description/?utm_source=LCUS&utm_medium=ip_redirect_q_uns&utm_campaign=transfer2china
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 1]: https://img-blog.csdnimg.cn/20191206104830370.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode.com_problems_binary-tree-paths]: https://leetcode.com/problems/binary-tree-paths/description/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 2]: https://img-blog.csdnimg.cn/20191206113106221.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 3]: https://img-blog.csdnimg.cn/20191206171103924.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 4]: https://img-blog.csdnimg.cn/20191206180015442.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode-cn.com_problems_combinations]: https://leetcode-cn.com/problems/combinations/description/?utm_source=LCUS&utm_medium=ip_redirect_q_uns&utm_campaign=transfer2china
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 5]: https://img-blog.csdnimg.cn/2019120710202313.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode.com_problems_combination-sum_description]: https://leetcode.com/problems/combination-sum/description/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 6]: https://img-blog.csdnimg.cn/20191207110956729.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode.com_problems_combination-sum-ii_description]: https://leetcode.com/problems/combination-sum-ii/description/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 7]: https://img-blog.csdnimg.cn/20191207143449937.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode.com_problems_combination-sum-iii_description]: https://leetcode.com/problems/combination-sum-iii/description/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 8]: https://img-blog.csdnimg.cn/20191207174333851.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode.com_problems_subsets-ii_description]: https://leetcode.com/problems/subsets-ii/description/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 9]: https://img-blog.csdnimg.cn/20191207175512337.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 10]: https://img-blog.csdnimg.cn/20191208121055159.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode-cn.com_problems_palindrome-partitioning]: https://leetcode-cn.com/problems/palindrome-partitioning/description/?utm_source=LCUS&utm_medium=ip_redirect_q_uns&utm_campaign=transfer2china
[20191208122902703.png]: https://img-blog.csdnimg.cn/20191208122902703.png
[https_leetcode-cn.com_problems_sudoku-solver_solution_jie-shu-du-by-leetcode]: https://leetcode-cn.com/problems/sudoku-solver/solution/jie-shu-du-by-leetcode/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 11]: https://img-blog.csdnimg.cn/20191208155709611.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 12]: https://img-blog.csdnimg.cn/20191208155809104.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[https_leetcode-cn.com_problems_n-queens_solution_nhuang-hou-by-leetcode]: https://leetcode-cn.com/problems/n-queens/solution/nhuang-hou-by-leetcode/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 13]: https://img-blog.csdnimg.cn/20191208195933354.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA_size_16_color_FFFFFF_t_70 14]: https://img-blog.csdnimg.cn/20191208200008967.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTkxMDY5NA==,size_16,color_FFFFFF,t_70

LeetCode——Backtracking

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