leetcode 51.N皇后

@(labuladong的算法小抄)[回溯]

leetcode 51.N皇后

题目描述

解题思路

参考https://labuladong.gitbook.io/algo/di-ling-zhang-bi-du-xi-lie/hui-su-suan-fa-xiang-jie-xiu-ding-ban

回溯

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

    public List<List<String>> solveNQueens(int n) {
        //记录路径
        char[][] track = new char[n][n];
        //初始化空棋盘
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                track[i][j] = '.';

        backtrack(track, 0);

        return res;
    }
    /*
    路径：小于row的所有行都已经存放在了track中
    选择列表：每一行的所有列都是选择
    结束条件：遍历完所有row，即track中已经存放了一张完整的满足条件的棋盘
     */
    private void backtrack(char[][] track, int row) {
        if (row == track.length) {
            //将track转变成ArrayList类型，添加进res中
            res.add(charToArrayList(track));
            return;
        }
        for (int col = 0; col < track.length; col++) {
            //如果当前列不能放置皇后，则直接跳过
            if (!isValid(track, row, col)) {
                continue;
            }
            //做出选择
            track[row][col] = 'Q';
            backtrack(track, row + 1);
            //撤销选择
            track[row][col] = '.';
        }
    }

    //将char[][]数组转变为ArrayList
    private List<String> charToArrayList(char[][] track) {
        List<String> afterChange = new ArrayList<>();
        for (char[] sub : track) {
            //把sub包装成String类型存放到afterChange中
            afterChange.add(String.valueOf(sub));
        }
        return afterChange;
    }

    //在当前棋盘track中，用于判断是否可以在[row,col]的位置放置皇后
    private boolean isValid(char[][] track, int row, int col) {
        boolean flag = true;
        //检查列是否有皇后冲突
        for (int i = row - 1; i >= 0; i--) {
            if (track[i][col] == 'Q')
                flag = false;
        }
        //检查左上方斜线是否有皇后冲突
        for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--){
            if (track[i][j] == 'Q')
                flag = false;
        }
        //检查右上方斜线是否有皇后冲突
        for (int i = row - 1, j = col + 1; i >=0 && j < track.length; i--, j++){
            if (track[i][j] == 'Q')
                flag = false;
        }
        return flag;
    }

}