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

时间:2023-03-09 07:48:10
[Leetcode][Python]52: N-Queens II
# -*- coding: utf8 -*-
'''
__author__ = 'dabay.wang@gmail.com' 52: N-Queens II
https://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)