代码随想录：回溯20-21

news2025/6/17 12:30:33

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res
        if(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String
                list.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Q
                backTracking(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;
    }

}