# -*- coding: utf8 -*-
'''
__author__ = 'dabay.wang@gmail.com' 51: N-Queens
https://oj.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. For example,
There exist two distinct solutions to the 4-queens puzzle:
[
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."], ["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
This set of questions is relatively simple , Start with a queen , Then find the next possible location and put the second one ... The trick is “ Find the next possible location ” On ,
- Next position , In fact, in the next line
- When checking if it can be placed , Just check if the column is occupied , And whether diagonal lines are occupied on the left and right respectively .（ Because there is no queen down there ）
'''
class Solution:
# @return a list of lists of string
def solveNQueens(self, n):
def make_solution(board):
copy = []
for row in board:
row_str = ""
for c in row:
row_str = row_str + c
copy.append(row_str)
return copy def check_up(r, c, queen_stack, board):
i = 1
while i < len(board):
if r-i>=0 and c-i>=0 and board[r-i][c-i]=='Q':
return False
if r-i>=0 and c+i<len(board) and board[r-i][c+i]=="Q":
return False
i = i + 1
else:
return True def find_available_positions(board, queen_stack):
positions = []
row = len(queen_stack)
queen_columns = [pos for pos in queen_stack]
for c in xrange(len(board)):
if c in queen_columns:
continue
if board[row][c] == "." and check_up(row, c, queen_stack, board):
positions.append((row,c))
return positions def DFS(board, queen_stack, res):
if len(queen_stack) >= len(board):
res.append(make_solution(board))
return
positions = find_available_positions(board, queen_stack)
for (r, c) in positions:
queen_stack.append((r, c))
board[r][c] = "Q"
DFS(board, queen_stack, res)
queen_stack.pop()
board[r][c] = "." board = [["."] * n for _ in xrange(n)]
queen_stack = []
res = [] DFS(board, queen_stack, res)
return res def print_board(board):
print '-' * 30
for row in board:
for item in row:
print item,
print
print '-' * 30 def main():
sol = Solution()
solutions = sol.solveNQueens(4)
for solution in solutions:
print_board(solution) if __name__ == "__main__":
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)

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