代码随想录：回溯20-21

51.N皇后

题目

按照国际象棋的规则，皇后可以攻击与之处在同一行或同一列或同一斜线上的棋子。

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。

给你一个整数 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。

示例 1：

输入：n = 4
输出：[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
解释：如上图所示，4 皇后问题存在两个不同的解法。

代码

class Solution {List<List<String>> res = new ArrayList<>();public List<List<String>> solveNQueens(int n) {//棋盘初始化全为.char[][] path = new char[n][n];for(char[] c : path){Arrays.fill(c,'.');}//从第一行开始回溯backTracking(n,0,path);return res;}public void backTracking(int n,int row,char[][] path){//row到了最后一行，把path转为List加入resif(row == n){List<String> list = new ArrayList<>();for(char[] c : path){ //获取棋盘的每一行char[]String s = String.copyValueOf(c); //把char[]转为Stringlist.add(s);}res.add(list); //把path对应的List加入结果集return;}//n个分支代表不同的列取值for(int col=0; col < n; col++){//如果此时path[row][col]的棋盘合法if(judge(row,col,path,n) == true){path[row][col] = 'Q';  //确定row行的QbackTracking(n,row+1,path);  //回溯row+1行path[row][col] = '.'; //回溯row行的Q，为下一个列做准备}}}public boolean judge(int row,int col,char[][] path,int n){//因为回溯过程中row一值变大，所以不用考虑同一行//同一列for(int i=row-1; i >=0; i--){if(path[i][col] == 'Q'){return false;}}//45斜对角for(int i=row-1,j=col-1; i>=0 && j>=0; i--,j--){if(path[i][j] == 'Q'){return false;}}//135斜对角for(int i=row-1,j=col+1; i>=0 && j<=n-1; i--,j++){if(path[i][j] == 'Q'){return false;}}return true;}
}

总结

树的每一层row代表棋盘的一行，里面的for循环代表棋盘的一列。row从0开始，当row到达n，说明走出棋盘了，作为叶子节点可以收集结果。

如果某个位置设置为Q,加入棋盘后导致棋盘不合法（同行、同列、同斜线），就要剪枝，但是剪枝是需要知道当前位置的row和col，所以把剪枝写在for循环里面，只有当前位置合法，才修改成Q，然后递归下一行row+1。如果不合法，直接跳过当前col，进入for循环的下一层循环。

判断位置是否合法，需要判断同列、同斜线，其中同斜线有两种情况，不要漏了。而且同一行不用判断，因为每调用一次backTracking，row都+1，相当于row一直向下走不会同一行。

37.解数独

题目

编写一个程序，通过填充空格来解决数独问题。

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。（请参考示例图）

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1：

输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：

代码

class Solution {public void solveSudoku(char[][] board) {backTracking(board);}public boolean backTracking(char[][] board){for(int i=0; i < 9; i++){for(int j=0; j < 9; j++){//已经填入了数字if(board[i][j] != '.'){continue;}for(char k = '1'; k <= '9'; k++){if(judge(i,j,k,board) == true){board[i][j] = k;//确定了位置[i][j]，递归下一个位置[i][j+1]//如果这里没有返回true，说明board[i][j] = k的条件下，下面是没有数独的解的if(backTracking(board)){return true;};board[i][j] = '.'; //下一个位置[i][j+1]找不到，说明这个k是错误解，要恢复成.,继续判断下一个数组k}}//53,53x的1-9都不行，说明数独无解，直接返回false，会终止return false;}}//两个for循环走完，还没返回，棋盘已经走完了，得到了唯一解return true; }public boolean judge(int row,int col,char k,char[][] board){//判断同一行for(int i = 0; i < 9;i++){if(board[row][i] == k){return false;}}//判断同一列for(int i = 0; i < 9;i++){if(board[i][col] == k){return false;}}//判断同一个九宫格int starti = row / 3 * 3;int startj = col / 3 * 3;for(int i = starti; i < starti + 3; i++){for(int j = startj; j < startj + 3; j++){if(board[i][j] == k){return false;}}}return true;}}