GuilinDev

Lc0051

05 August 2008

51 N-Queens

原题概述

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens’ placement, where 

1

'Q'
and 
1

'.'
both indicate a queen and an empty space respectively.

Example:

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Input: 4
Output: [
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

题意和分析

N皇后问题，棋盘上每一行每一列和每一斜线都不能都两个皇后；这道题是经典的回溯法，按照每一行遍历，每次加一个皇后，然后利用回溯法检查每一列和每个对角线是否已经有皇后了，有则回溯到前面一个状态，否则继续向下递归，直到最后一行的皇后放好就将结果加入到results里面。时间复杂度O(n^2)，空间会创建一个记录每一行皇后所在的列的位置的一维数组，为O(n)。

代码

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class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> results = new ArrayList<>();
        if (n <= 0) {
            return results;
        }
        // int[n]是记录每一行皇后所放的位置
        // int[0]是第一行皇后的位置，int[1]是第二行皇后的位置...
        dfs(results, 0, new int[n]);
        return results;
    }
    //传入的参数是结果集，循环的行数，循环到该行-皇后所处的列的位置
    private void dfs(List<List<String>> result, int index, int[] queenPositions) {
        //递归的结束条件，找到一个结果
        if (index == queenPositions.length) {
            addOneSolution(result, queenPositions);
            return;
        }

        // 逐行构建一个有效的放置方法
        for (int i = 0; i < queenPositions.length; i++) {
            queenPositions[index] = i; //每个位置都尝试，找到所有可能的组合
            if (isValid(queenPositions, index)) {
                dfs(result, index + 1, queenPositions);
            }
        }
    }

    private boolean isValid(int[] queenPositions, int index) {
        for (int i = 0; i < index; i++) {
            //剪枝，判断同一列中是否有重复的皇后
            if (queenPositions[i] == queenPositions[index]) {
                return false;
            }
            //判断对角线上是否有重复的皇后
            if (Math.abs(queenPositions[index] - queenPositions[i]) == Math.abs(i - index)) {
                return false;
            }
        }
        return true;
    }

    // 根据每行皇后的位置构建结果并返回
    private void addOneSolution(List<List<String>> result, int[] queens) {
        List<String> oneResult = new ArrayList<>();
        for (int queen : queens) {
            StringBuilder sb = new StringBuilder();
            for (int j = 0; j < queens.length; j++) {
                if (queen == j) {
                    sb.append('Q');
                } else {
                    sb.append('.');
                }
            }
            oneResult.add(sb.toString());
        }
        result.add(oneResult);
    }
}
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