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867 lines
24 KiB
Markdown
867 lines
24 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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# 51. N皇后
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[力扣题目链接](https://leetcode.cn/problems/n-queens/)
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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。
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给你一个整数 n ，返回所有不同的 n 皇后问题 的解决方案。
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每一种解法包含一个不同的 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。
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示例 1：
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* 输入：n = 4
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* 输出：[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
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* 解释：如上图所示，4 皇后问题存在两个不同的解法。
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示例 2：
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* 输入：n = 1
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* 输出：[["Q"]]
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# 算法公开课
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**《代码随想录》算法视频公开课：[这就是传说中的N皇后？ 回溯算法安排！| LeetCode：51.N皇后](https://www.bilibili.com/video/BV1Rd4y1c7Bq/)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
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## 思路
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都知道n皇后问题是回溯算法解决的经典问题，但是用回溯解决多了组合、切割、子集、排列问题之后，遇到这种二维矩阵还会有点不知所措。
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首先来看一下皇后们的约束条件：
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1. 不能同行
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2. 不能同列
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3. 不能同斜线
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确定完约束条件，来看看究竟要怎么去搜索皇后们的位置，其实搜索皇后的位置，可以抽象为一棵树。
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下面我用一个 3 * 3 的棋盘，将搜索过程抽象为一棵树，如图：
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从图中，可以看出，二维矩阵中矩阵的高就是这棵树的高度，矩阵的宽就是树形结构中每一个节点的宽度。
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那么我们用皇后们的约束条件，来回溯搜索这棵树，**只要搜索到了树的叶子节点，说明就找到了皇后们的合理位置了**。
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### 回溯三部曲
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按照我总结的如下回溯模板，我们来依次分析：
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```
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void backtracking(参数) {
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if (终止条件) {
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存放结果;
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return;
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}
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for (选择：本层集合中元素（树中节点孩子的数量就是集合的大小）) {
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处理节点;
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backtracking(路径，选择列表); // 递归
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回溯，撤销处理结果
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}
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}
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```
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* 递归函数参数
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我依然是定义全局变量二维数组result来记录最终结果。
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参数n是棋盘的大小，然后用row来记录当前遍历到棋盘的第几层了。
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代码如下：
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```cpp
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vector<vector<string>> result;
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void backtracking(int n, int row, vector<string>& chessboard) {
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```
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* 递归终止条件
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在如下树形结构中：
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可以看出，当递归到棋盘最底层（也就是叶子节点）的时候，就可以收集结果并返回了。
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代码如下：
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```cpp
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if (row == n) {
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result.push_back(chessboard);
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return;
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}
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```
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* 单层搜索的逻辑
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递归深度就是row控制棋盘的行，每一层里for循环的col控制棋盘的列，一行一列，确定了放置皇后的位置。
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每次都是要从新的一行的起始位置开始搜，所以都是从0开始。
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代码如下：
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```cpp
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for (int col = 0; col < n; col++) {
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if (isValid(row, col, chessboard, n)) { // 验证合法就可以放
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chessboard[row][col] = 'Q'; // 放置皇后
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backtracking(n, row + 1, chessboard);
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chessboard[row][col] = '.'; // 回溯，撤销皇后
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}
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}
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```
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* 验证棋盘是否合法
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按照如下标准去重：
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1. 不能同行
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2. 不能同列
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3. 不能同斜线 （45度和135度角）
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代码如下：
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```CPP
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bool isValid(int row, int col, vector<string>& chessboard, int n) {
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// 检查列
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for (int i = 0; i < row; i++) { // 这是一个剪枝
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if (chessboard[i][col] == 'Q') {
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return false;
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}
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}
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// 检查 45度角是否有皇后
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for (int i = row - 1, j = col - 1; i >=0 && j >= 0; i--, j--) {
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if (chessboard[i][j] == 'Q') {
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return false;
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}
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}
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// 检查 135度角是否有皇后
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for(int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {
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if (chessboard[i][j] == 'Q') {
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return false;
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}
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}
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return true;
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}
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```
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在这份代码中，细心的同学可以发现为什么没有在同行进行检查呢？
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因为在单层搜索的过程中，每一层递归，只会选for循环（也就是同一行）里的一个元素，所以不用去重了。
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那么按照这个模板不难写出如下C++代码：
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```CPP
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class Solution {
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private:
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vector<vector<string>> result;
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// n 为输入的棋盘大小
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// row 是当前递归到棋盘的第几行了
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void backtracking(int n, int row, vector<string>& chessboard) {
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if (row == n) {
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result.push_back(chessboard);
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return;
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}
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for (int col = 0; col < n; col++) {
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if (isValid(row, col, chessboard, n)) { // 验证合法就可以放
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chessboard[row][col] = 'Q'; // 放置皇后
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backtracking(n, row + 1, chessboard);
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chessboard[row][col] = '.'; // 回溯，撤销皇后
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}
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}
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}
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bool isValid(int row, int col, vector<string>& chessboard, int n) {
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// 检查列
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for (int i = 0; i < row; i++) { // 这是一个剪枝
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if (chessboard[i][col] == 'Q') {
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return false;
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}
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}
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// 检查 45度角是否有皇后
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for (int i = row - 1, j = col - 1; i >=0 && j >= 0; i--, j--) {
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if (chessboard[i][j] == 'Q') {
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return false;
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}
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}
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// 检查 135度角是否有皇后
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for(int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {
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if (chessboard[i][j] == 'Q') {
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return false;
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}
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}
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return true;
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}
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public:
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vector<vector<string>> solveNQueens(int n) {
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result.clear();
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std::vector<std::string> chessboard(n, std::string(n, '.'));
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backtracking(n, 0, chessboard);
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return result;
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}
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};
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```
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* 时间复杂度: O(n!)
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* 空间复杂度: O(n)
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可以看出，除了验证棋盘合法性的代码，省下来部分就是按照回溯法模板来的。
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## 总结
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本题是我们解决棋盘问题的第一道题目。
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如果从来没有接触过N皇后问题的同学看着这样的题会感觉无从下手，可能知道要用回溯法，但也不知道该怎么去搜。
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**这里我明确给出了棋盘的宽度就是for循环的长度，递归的深度就是棋盘的高度，这样就可以套进回溯法的模板里了**。
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大家可以在仔细体会体会！
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## 其他语言补充
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### Java
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```java
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class Solution {
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List<List<String>> res = new ArrayList<>();
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public List<List<String>> solveNQueens(int n) {
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char[][] chessboard = new char[n][n];
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for (char[] c : chessboard) {
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Arrays.fill(c, '.');
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}
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backTrack(n, 0, chessboard);
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return res;
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}
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public void backTrack(int n, int row, char[][] chessboard) {
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if (row == n) {
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res.add(Array2List(chessboard));
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return;
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}
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for (int col = 0;col < n; ++col) {
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if (isValid (row, col, n, chessboard)) {
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chessboard[row][col] = 'Q';
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backTrack(n, row+1, chessboard);
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chessboard[row][col] = '.';
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}
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}
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}
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public List Array2List(char[][] chessboard) {
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List<String> list = new ArrayList<>();
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for (char[] c : chessboard) {
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list.add(String.copyValueOf(c));
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}
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return list;
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}
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public boolean isValid(int row, int col, int n, char[][] chessboard) {
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// 检查列
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for (int i=0; i<row; ++i) { // 相当于剪枝
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if (chessboard[i][col] == 'Q') {
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return false;
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}
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}
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// 检查45度对角线
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for (int i=row-1, j=col-1; i>=0 && j>=0; i--, j--) {
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if (chessboard[i][j] == 'Q') {
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return false;
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}
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}
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// 检查135度对角线
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for (int i=row-1, j=col+1; i>=0 && j<=n-1; i--, j++) {
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if (chessboard[i][j] == 'Q') {
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return false;
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}
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}
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return true;
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}
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}
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```
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```java
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// 方法2：使用boolean数组表示已经占用的直(斜)线
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class Solution {
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List<List<String>> res = new ArrayList<>();
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boolean[] usedCol, usedDiag45, usedDiag135; // boolean数组中的每个元素代表一条直(斜)线
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public List<List<String>> solveNQueens(int n) {
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usedCol = new boolean[n]; // 列方向的直线条数为 n
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usedDiag45 = new boolean[2 * n - 1]; // 45°方向的斜线条数为 2 * n - 1
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usedDiag135 = new boolean[2 * n - 1]; // 135°方向的斜线条数为 2 * n - 1
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//用于收集结果, 元素的index表示棋盘的row，元素的value代表棋盘的column
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int[] board = new int[n];
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backTracking(board, n, 0);
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return res;
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}
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private void backTracking(int[] board, int n, int row) {
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if (row == n) {
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//收集结果
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List<String> temp = new ArrayList<>();
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for (int i : board) {
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char[] str = new char[n];
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Arrays.fill(str, '.');
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str[i] = 'Q';
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temp.add(new String(str));
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}
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res.add(temp);
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return;
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}
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for (int col = 0; col < n; col++) {
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if (usedCol[col] | usedDiag45[row + col] | usedDiag135[row - col + n - 1]) {
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continue;
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}
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board[row] = col;
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// 标记该列出现过
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usedCol[col] = true;
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// 同一45°斜线上元素的row + col为定值, 且各不相同
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usedDiag45[row + col] = true;
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// 同一135°斜线上元素row - col为定值, 且各不相同
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// row - col 值有正有负, 加 n - 1 是为了对齐零点
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usedDiag135[row - col + n - 1] = true;
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// 递归
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backTracking(board, n, row + 1);
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usedCol[col] = false;
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usedDiag45[row + col] = false;
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usedDiag135[row - col + n - 1] = false;
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}
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}
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}
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```
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### Python
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```python
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class Solution:
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def solveNQueens(self, n: int) -> List[List[str]]:
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result = [] # 存储最终结果的二维字符串数组
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chessboard = ['.' * n for _ in range(n)] # 初始化棋盘
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self.backtracking(n, 0, chessboard, result) # 回溯求解
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return [[''.join(row) for row in solution] for solution in result] # 返回结果集
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def backtracking(self, n: int, row: int, chessboard: List[str], result: List[List[str]]) -> None:
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if row == n:
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result.append(chessboard[:]) # 棋盘填满，将当前解加入结果集
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return
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for col in range(n):
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if self.isValid(row, col, chessboard):
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chessboard[row] = chessboard[row][:col] + 'Q' + chessboard[row][col+1:] # 放置皇后
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self.backtracking(n, row + 1, chessboard, result) # 递归到下一行
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chessboard[row] = chessboard[row][:col] + '.' + chessboard[row][col+1:] # 回溯，撤销当前位置的皇后
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def isValid(self, row: int, col: int, chessboard: List[str]) -> bool:
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# 检查列
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for i in range(row):
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if chessboard[i][col] == 'Q':
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return False # 当前列已经存在皇后，不合法
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# 检查 45 度角是否有皇后
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i, j = row - 1, col - 1
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while i >= 0 and j >= 0:
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if chessboard[i][j] == 'Q':
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return False # 左上方向已经存在皇后，不合法
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i -= 1
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j -= 1
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# 检查 135 度角是否有皇后
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i, j = row - 1, col + 1
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while i >= 0 and j < len(chessboard):
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if chessboard[i][j] == 'Q':
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return False # 右上方向已经存在皇后，不合法
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i -= 1
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j += 1
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return True # 当前位置合法
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```
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### Go
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```Go
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func solveNQueens(n int) [][]string {
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var res [][]string
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chessboard := make([][]string, n)
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for i := 0; i < n; i++ {
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chessboard[i] = make([]string, n)
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}
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for i := 0; i < n; i++ {
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for j := 0; j < n; j++ {
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chessboard[i][j] = "."
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}
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}
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var backtrack func(int)
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backtrack = func(row int) {
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if row == n {
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temp := make([]string, n)
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for i, rowStr := range chessboard {
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temp[i] = strings.Join(rowStr, "")
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}
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res = append(res, temp)
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return
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}
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for i := 0; i < n; i++ {
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if isValid(n, row, i, chessboard) {
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chessboard[row][i] = "Q"
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backtrack(row + 1)
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chessboard[row][i] = "."
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}
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}
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}
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backtrack(0)
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return res
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}
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func isValid(n, row, col int, chessboard [][]string) bool {
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for i := 0; i < row; i++ {
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if chessboard[i][col] == "Q" {
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return false
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}
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}
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for i, j := row-1, col-1; i >= 0 && j >= 0; i, j = i-1, j-1 {
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if chessboard[i][j] == "Q" {
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return false
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}
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}
|
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for i, j := row-1, col+1; i >= 0 && j < n; i, j = i-1, j+1 {
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||
if chessboard[i][j] == "Q" {
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return false
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}
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}
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return true
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}
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```
|
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|
||
|
||
### Javascript
|
||
```Javascript
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var solveNQueens = function(n) {
|
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function isValid(row, col, chessBoard, n) {
|
||
|
||
for(let i = 0; i < row; i++) {
|
||
if(chessBoard[i][col] === 'Q') {
|
||
return false
|
||
}
|
||
}
|
||
|
||
for(let i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {
|
||
if(chessBoard[i][j] === 'Q') {
|
||
return false
|
||
}
|
||
}
|
||
|
||
for(let i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {
|
||
if(chessBoard[i][j] === 'Q') {
|
||
return false
|
||
}
|
||
}
|
||
return true
|
||
}
|
||
|
||
function transformChessBoard(chessBoard) {
|
||
let chessBoardBack = []
|
||
chessBoard.forEach(row => {
|
||
let rowStr = ''
|
||
row.forEach(value => {
|
||
rowStr += value
|
||
})
|
||
chessBoardBack.push(rowStr)
|
||
})
|
||
|
||
return chessBoardBack
|
||
}
|
||
|
||
let result = []
|
||
function backtracing(row,chessBoard) {
|
||
if(row === n) {
|
||
result.push(transformChessBoard(chessBoard))
|
||
return
|
||
}
|
||
for(let col = 0; col < n; col++) {
|
||
if(isValid(row, col, chessBoard, n)) {
|
||
chessBoard[row][col] = 'Q'
|
||
backtracing(row + 1,chessBoard)
|
||
chessBoard[row][col] = '.'
|
||
}
|
||
}
|
||
}
|
||
let chessBoard = new Array(n).fill([]).map(() => new Array(n).fill('.'))
|
||
backtracing(0,chessBoard)
|
||
return result
|
||
|
||
};
|
||
```
|
||
|
||
### TypeScript
|
||
|
||
```typescript
|
||
function solveNQueens(n: number): string[][] {
|
||
const board: string[][] = new Array(n).fill(0).map(_ => new Array(n).fill('.'));
|
||
const resArr: string[][] = [];
|
||
backTracking(n, 0, board);
|
||
return resArr;
|
||
function backTracking(n: number, rowNum: number, board: string[][]): void {
|
||
if (rowNum === n) {
|
||
resArr.push(transformBoard(board));
|
||
return;
|
||
}
|
||
for (let i = 0; i < n; i++) {
|
||
if (isValid(i, rowNum, board) === true) {
|
||
board[rowNum][i] = 'Q';
|
||
backTracking(n, rowNum + 1, board);
|
||
board[rowNum][i] = '.';
|
||
}
|
||
}
|
||
}
|
||
};
|
||
function isValid(col: number, row: number, board: string[][]): boolean {
|
||
const n: number = board.length;
|
||
if (col < 0 || col >= n || row < 0 || row >= n) return false;
|
||
// 检查列
|
||
for (let row of board) {
|
||
if (row[col] === 'Q') return false;
|
||
}
|
||
// 检查45度方向
|
||
let x: number = col,
|
||
y: number = row;
|
||
while (y >= 0 && x < n) {
|
||
if (board[y--][x++] === 'Q') return false;
|
||
}
|
||
// 检查135度方向
|
||
x = col;
|
||
y = row;
|
||
while (x >= 0 && y >= 0) {
|
||
if (board[y--][x--] === 'Q') return false;
|
||
}
|
||
return true;
|
||
}
|
||
function transformBoard(board: string[][]): string[] {
|
||
const resArr = [];
|
||
for (let row of board) {
|
||
resArr.push(row.join(''));
|
||
}
|
||
return resArr;
|
||
}
|
||
```
|
||
|
||
### Swift
|
||
|
||
```swift
|
||
func solveNQueens(_ n: Int) -> [[String]] {
|
||
var result = [[String]]()
|
||
// 棋盘，使用Character的二维数组，以便于更新元素
|
||
var chessboard = [[Character]](repeating: [Character](repeating: ".", count: n), count: n)
|
||
// 检查棋盘是否符合N皇后
|
||
func isVaild(row: Int, col: Int) -> Bool {
|
||
// 检查列
|
||
for i in 0 ..< row {
|
||
if chessboard[i][col] == "Q" { return false }
|
||
}
|
||
|
||
var i, j: Int
|
||
// 检查45度
|
||
i = row - 1
|
||
j = col - 1
|
||
while i >= 0, j >= 0 {
|
||
if chessboard[i][j] == "Q" { return false }
|
||
i -= 1
|
||
j -= 1
|
||
}
|
||
// 检查135度
|
||
i = row - 1
|
||
j = col + 1
|
||
while i >= 0, j < n {
|
||
if chessboard[i][j] == "Q" { return false }
|
||
i -= 1
|
||
j += 1
|
||
}
|
||
|
||
return true
|
||
}
|
||
func backtracking(row: Int) {
|
||
if row == n {
|
||
result.append(chessboard.map { String($0) })
|
||
}
|
||
|
||
for col in 0 ..< n {
|
||
guard isVaild(row: row, col: col) else { continue }
|
||
chessboard[row][col] = "Q" // 放置皇后
|
||
backtracking(row: row + 1)
|
||
chessboard[row][col] = "." // 回溯
|
||
}
|
||
}
|
||
backtracking(row: 0)
|
||
return result
|
||
}
|
||
```
|
||
|
||
### Rust
|
||
|
||
```Rust
|
||
impl Solution {
|
||
fn is_valid(row: usize, col: usize, chessboard: &mut Vec<Vec<char>>, n: usize) -> bool {
|
||
let mut i = 0 as usize;
|
||
while i < row {
|
||
if chessboard[i][col] == 'Q' { return false; }
|
||
i += 1;
|
||
}
|
||
let (mut i, mut j) = (row as i32 - 1, col as i32 - 1);
|
||
while i >= 0 && j >= 0 {
|
||
if chessboard[i as usize][j as usize] == 'Q' { return false; }
|
||
i -= 1;
|
||
j -= 1;
|
||
}
|
||
let (mut i, mut j) = (row as i32 - 1, col as i32 + 1);
|
||
while i >= 0 && j < n as i32 {
|
||
if chessboard[i as usize][j as usize] == 'Q' { return false; }
|
||
i -= 1;
|
||
j += 1;
|
||
}
|
||
return true;
|
||
}
|
||
fn backtracking(result: &mut Vec<Vec<String>>, n: usize, row: usize, chessboard: &mut Vec<Vec<char>>) {
|
||
if row == n {
|
||
let mut chessboard_clone: Vec<String> = Vec::new();
|
||
for i in chessboard {
|
||
chessboard_clone.push(i.iter().collect::<String>());
|
||
}
|
||
result.push(chessboard_clone);
|
||
return;
|
||
}
|
||
for col in 0..n {
|
||
if Self::is_valid(row, col, chessboard, n) {
|
||
chessboard[row][col] = 'Q';
|
||
Self::backtracking(result, n, row + 1, chessboard);
|
||
chessboard[row][col] = '.';
|
||
}
|
||
}
|
||
}
|
||
pub fn solve_n_queens(n: i32) -> Vec<Vec<String>> {
|
||
let mut result: Vec<Vec<String>> = Vec::new();
|
||
let mut chessboard: Vec<Vec<char>> = vec![vec!['.'; n as usize]; n as usize];
|
||
Self::backtracking(&mut result, n as usize, 0, &mut chessboard);
|
||
result
|
||
}
|
||
}
|
||
```
|
||
|
||
### C
|
||
```c
|
||
char ***ans;
|
||
char **path;
|
||
int ansTop, pathTop;
|
||
//将path中元素复制到ans中
|
||
void copyPath(int n) {
|
||
char **tempPath = (char**)malloc(sizeof(char*) * pathTop);
|
||
int i;
|
||
for(i = 0; i < pathTop; ++i) {
|
||
tempPath[i] = (char*)malloc(sizeof(char) * n + 1);
|
||
int j;
|
||
for(j = 0; j < n; ++j)
|
||
tempPath[i][j] = path[i][j];
|
||
tempPath[i][j] = '\0';
|
||
|
||
}
|
||
ans[ansTop++] = tempPath;
|
||
}
|
||
|
||
//判断当前位置是否有效（是否不被其它皇后影响）
|
||
int isValid(int x, int y, int n) {
|
||
int i, j;
|
||
//检查同一行以及同一列是否有效
|
||
for(i = 0; i < n; ++i) {
|
||
if(path[y][i] == 'Q' || path[i][x] == 'Q')
|
||
return 0;
|
||
}
|
||
//下面两个for循环检查斜角45度是否有效
|
||
i = y - 1;
|
||
j = x - 1;
|
||
while(i >= 0 && j >= 0) {
|
||
if(path[i][j] == 'Q')
|
||
return 0;
|
||
--i, --j;
|
||
}
|
||
|
||
i = y + 1;
|
||
j = x + 1;
|
||
while(i < n && j < n) {
|
||
if(path[i][j] == 'Q')
|
||
return 0;
|
||
++i, ++j;
|
||
}
|
||
|
||
//下面两个for循环检查135度是否有效
|
||
i = y - 1;
|
||
j = x + 1;
|
||
while(i >= 0 && j < n) {
|
||
if(path[i][j] == 'Q')
|
||
return 0;
|
||
--i, ++j;
|
||
}
|
||
|
||
i = y + 1;
|
||
j = x -1;
|
||
while(j >= 0 && i < n) {
|
||
if(path[i][j] == 'Q')
|
||
return 0;
|
||
++i, --j;
|
||
}
|
||
return 1;
|
||
}
|
||
|
||
void backTracking(int n, int depth) {
|
||
//若path中有四个元素，将其拷贝到ans中。从当前层返回
|
||
if(pathTop == n) {
|
||
copyPath(n);
|
||
return;
|
||
}
|
||
|
||
//遍历横向棋盘
|
||
int i;
|
||
for(i = 0; i < n; ++i) {
|
||
//若当前位置有效
|
||
if(isValid(i, depth, n)) {
|
||
//在当前位置放置皇后
|
||
path[depth][i] = 'Q';
|
||
//path中元素数量+1
|
||
++pathTop;
|
||
|
||
backTracking(n, depth + 1);
|
||
//进行回溯
|
||
path[depth][i] = '.';
|
||
//path中元素数量-1
|
||
--pathTop;
|
||
}
|
||
}
|
||
}
|
||
|
||
//初始化存储char*数组path，将path中所有元素设为'.'
|
||
void initPath(int n) {
|
||
int i, j;
|
||
for(i = 0; i < n; i++) {
|
||
//为path中每个char*开辟空间
|
||
path[i] = (char*)malloc(sizeof(char) * n + 1);
|
||
//将path中所有字符设为'.'
|
||
for(j = 0; j < n; j++)
|
||
path[i][j] = '.';
|
||
//在每个字符串结尾加入'\0'
|
||
path[i][j] = '\0';
|
||
}
|
||
}
|
||
|
||
char *** solveNQueens(int n, int* returnSize, int** returnColumnSizes){
|
||
//初始化辅助变量
|
||
ans = (char***)malloc(sizeof(char**) * 400);
|
||
path = (char**)malloc(sizeof(char*) * n);
|
||
ansTop = pathTop = 0;
|
||
|
||
//初始化path数组
|
||
initPath(n);
|
||
backTracking(n, 0);
|
||
|
||
//设置返回数组大小
|
||
*returnSize = ansTop;
|
||
int i;
|
||
*returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
|
||
for(i = 0; i < ansTop; ++i) {
|
||
(*returnColumnSizes)[i] = n;
|
||
}
|
||
return ans;
|
||
}
|
||
```
|
||
|
||
### Scala
|
||
|
||
```scala
|
||
object Solution {
|
||
import scala.collection.mutable
|
||
def solveNQueens(n: Int): List[List[String]] = {
|
||
var result = mutable.ListBuffer[List[String]]()
|
||
|
||
def judge(x: Int, y: Int, maze: Array[Array[Boolean]]): Boolean = {
|
||
// 正上方
|
||
var xx = x
|
||
while (xx >= 0) {
|
||
if (maze(xx)(y)) return false
|
||
xx -= 1
|
||
}
|
||
// 左边
|
||
var yy = y
|
||
while (yy >= 0) {
|
||
if (maze(x)(yy)) return false
|
||
yy -= 1
|
||
}
|
||
// 左上方
|
||
xx = x
|
||
yy = y
|
||
while (xx >= 0 && yy >= 0) {
|
||
if (maze(xx)(yy)) return false
|
||
xx -= 1
|
||
yy -= 1
|
||
}
|
||
xx = x
|
||
yy = y
|
||
// 右上方
|
||
while (xx >= 0 && yy < n) {
|
||
if (maze(xx)(yy)) return false
|
||
xx -= 1
|
||
yy += 1
|
||
}
|
||
true
|
||
}
|
||
|
||
def backtracking(row: Int, maze: Array[Array[Boolean]]): Unit = {
|
||
if (row == n) {
|
||
// 将结果转换为题目所需要的形式
|
||
var path = mutable.ListBuffer[String]()
|
||
for (x <- maze) {
|
||
var tmp = mutable.ListBuffer[String]()
|
||
for (y <- x) {
|
||
if (y == true) tmp.append("Q")
|
||
else tmp.append(".")
|
||
}
|
||
path.append(tmp.mkString)
|
||
}
|
||
result.append(path.toList)
|
||
return
|
||
}
|
||
|
||
for (j <- 0 until n) {
|
||
// 判断这个位置是否可以放置皇后
|
||
if (judge(row, j, maze)) {
|
||
maze(row)(j) = true
|
||
backtracking(row + 1, maze)
|
||
maze(row)(j) = false
|
||
}
|
||
}
|
||
}
|
||
|
||
backtracking(0, Array.ofDim[Boolean](n, n))
|
||
result.toList
|
||
}
|
||
}
|
||
```
|
||
|
||
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|