[Leetcode][Python]52: N-Queens II

# -*- coding: utf8 -*-‘‘‘__author__ = ‘[email protected]‘

52: N-Queens IIhttps://oj.leetcode.com/problems/n-queens-ii/

Follow up for N-Queens problem.Now, instead outputting board configurations, return the total number of distinct solutions.

===Comments by Dabay===不知道和N Queen相比有没有简单很多的方法。我这里的解法思路和N Queen一样的。

一个一个放皇后，知道能放下最后一个皇后，解法+1。放第k个皇后的时候，在第k行中找位置，先看列被占用没有，然后往左上和右上看斜线被占用没有。‘‘‘

class Solution:    # @return an integer    def totalNQueens(self, n):        def check_up(r, c, board):            for row in xrange(r):                if board[row][c] == ‘Q‘:                    return False            else:                return True

        def check_upleft(r, c, board):            row = r - 1            column = c - 1            while row>=0 and column>=0:                if board[row][column] == ‘Q‘:                    return False                row = row - 1                column = column - 1            else:                return True

        def check_upright(r, c, board):            row = r - 1            column = c + 1            while row>=0 and column<len(board):                if board[row][column] == ‘Q‘:                    return False                row = row - 1                column = column + 1            else:                return True

        def DFS(board, queens, res):            if queens == 0:                res[0] = res[0] + 1                return            r = len(board) - queens            for c in xrange(len(board)):                if not check_up(r, c, board) or not check_upleft(r, c, board) or not check_upright(r, c, board):                    continue                else:                    board[r][c] = ‘Q‘                    DFS(board, queens-1, res)                    board[r][c] = ‘.‘

        board = [[‘.‘] * n for _ in xrange(n)]        #print board        queens = n        res = [0]        DFS(board, queens, res)        return res[0]

def main():    sol = Solution()    print sol.totalNQueens(5)

if __name__ == "__main__":    import time    start = time.clock()    main()    print "%s sec" % (time.clock() - start)