[LeetCode]题解（python）：051-N-Queens

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题目来源

https://leetcode.com/problems/n-queens/

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

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题意分析
Input: n:integer

Output:a list contains the result

Conditions:典型的n皇后问题（同一列，同一行，同斜线会互相攻击）

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题目思路

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本题很明显用回溯的方法，经大神指点，这里采用一维数组的方式，如n=4时，list = [2，4，1，3]表示i行第list[i]位置放置皇后，这样就保证了每一行只有一位皇后，然后代码中有个check条件，就是用来判断是不是同一列和斜线的，

注意如果list[i] == list[j]则表示同一列，abs（i - j） == abs（list[i] - list[j]）表示同斜线（很明显同一斜线时两点的横纵坐标差相等）

之后，回溯就好了

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AC代码（Python）

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__author__ = 'YE'

class Solution(object):
    def solveNQueens(self, n):
        """
        :type n: int
        :rtype: List[List[str]]
        """
        def check(k, j):
            for i in range(k):
                if board[i] == j or abs(k - i) == abs(board[i] - j):
                    return False
            return True

        def dfs(depth, valueList):
            if depth == n:
                res.append(valueList)
            else:
                for i in range(n):
                    if check(depth, i):
                        board[depth] = i
                        s = '.' * n
                        dfs(depth + 1, valueList + [s[:i] + "Q" + s[i+1:]])
        res = []

        board = [-1 for i in range(n)]
        dfs(0, [])

        return res

s = Solution()
n = 4
print(s.solveNQueens(n))