LC51— N 皇后

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。

输入：n = 1
输出：[["Q"]]

package com.nie.o3;/* * *@auth wenzhao *@date 2021/3/15 14:59 */


import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;

public class LEE51 {
   
    List<List<String>> res = new ArrayList<>();

    public List<List<String>> solveNQueens(int n) {
   
        char[][] board = new char[n][n];
        for (char[] i : board) {
   
            Arrays.fill(i, '.');
        }
        backrack(board, 0);
        return res;
    }

    /** * 路径: board 中小于row的那些行都已经成功的防止了皇后 * 选择列表: 第row行的所有列都是放置皇后的选择 * 结束条件: row超过了board的最后一行 * * @param board */
    private void backrack(char[][] board, int row) {
   
        if (row == board.length) {
   
            res.add(array2List(board));
            return;
        }

        for (int i = 0; i < board.length; i++) {
   
            if (!check(board, row, i)) {
   
                continue;
            }
            board[row][i] = 'Q';
            backrack(board, row + 1);
            board[row][i] = '.';
        }
    }

    private boolean check(char[][] board, int row, int col) {
   
        int n = board.length;
        //检查是否有皇后冲突 相互
        for (int i = 0; i < n; i++) {
   
            if (board[i][col] == 'Q')
                return false;
        }

        //检查右上方,是否有皇后相互冲突
        for (int i = row - 1, j = col + 1; i >= 0 && j < n;
             i--, j++) {
   
            if (board[i][j] == 'Q')
                return false;
        }

        //检查左上方,是否有皇后相互冲突
        for (int i = row - 1, j = col - 1; i >= 0 && j >= 0;
             i--, j--) {
   
            if (board[i][j] == 'Q')
                return false;
        }
        return true;
    }

    private List<String> array2List(char[][] board) {
   
        List<String> res = new LinkedList<>();
        for (char[] i : board) {
   
            StringBuffer sb = new StringBuffer();
            for (char j : i) {
   
                sb.append(j);
            }
            res.add(sb.toString());
        }
        return res;
    }


}