原题传送门

1. 题目描述

2. Solution

1、思路分析
遍历每一个空位，检查当前位置是否可以填入'Q' (即检查列、45度对角线、135度负对角线)，如合法则在当前位置填入'Q'，

2、代码实现

public class Solution {

    public List<List<String>> solveNQueens(int n) {
        char[][] chess = new char[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                chess[i][j] = '.';
            }
        }
        List<List<String>> res = new ArrayList<>();

        solve(res, chess, 0);
        return res;
    }

    private void solve(List<List<String>> res, char[][] chess, int row) {
        if (row == chess.length) {
            res.add(construct(chess));
            return;
        }
        for (int col = 0; col < chess.length; col++) {
            if (valid(chess, row, col)) {
                chess[row][col] = 'Q';
                solve(res, chess, row + 1);
                chess[row][col] = '.';
            }
        }

    }

    private boolean valid(char[][] chess, int row, int col) {
        // check all cols
        for (int i = 0; i < row; i++) {
            if (chess[i][col] == 'Q') return false;
        }

        // check 45 degree
        for (int i = row - 1, j = col + 1; i >= 0 && j < chess.length; i--, j++) {
            if (chess[i][j] == 'Q') return false;
        }

        // check 135 degree
        for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {
            if (chess[i][j] == 'Q') return false;
        }

        return true;
    }

    private List<String> construct(char[][] chess) {
        List<String> path = new ArrayList<>();
        for (char[] chars : chess)
            path.add(new String(chars));
        return path;
    }
}

3、复杂度分析
时间复杂度: O(n!)
空间复杂度: O(n^2)

标签：return,int,0051,LeetCode,char,chess,Queens,col,row
来源： https://www.cnblogs.com/junstat/p/16090018.html