51. 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.
image
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.
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.
My Solution:
import copy
class Solution:
def solveNQueens(self, n):
"""
:type n: int
:rtype: List[List[str]]
"""
result = []
board = [[0 * m for m in range(n)] for m in range(n)]
row = 0
self.get_board(n, row, result, board)
for r in result:
for i in range(len(r)):
r[i] = ''.join(str(s) for s in r[i])
return result
def get_board(self, n, row, result, board):
if row >= n:
result.append(board)
return
for column in range(n):
if board[row][column] == 0:
board, temp_board = self.new_board(n, row, column, board)
self.get_board(n, row+1, result, board)
board = temp_board.copy()
def new_board(self, n, x, y, board):
temp_board = copy.deepcopy(board)
# 方向数组，对于8个方向
board[x][y] = 'Q'
dx = [0, 0, -1, 1, -1, 1, -1, 1] # 上 下 左 右 左上 右上 左下 右下
dy = [-1, 1, 0, 0, -1, -1, 1, 1] # 上 下 左 右 左上 右上 左下 右下
for i in range(8):
for j in range(1, n):
new_x = x + dx[i] * j
new_y = y + dy[i] * j
if (new_x >= 0 and new_x < n) and (new_y >= 0 and new_y < n):
board[new_x][new_y] = '.'
return board, temp_board
Reference:
class Solution:
def solveNQueens(self, n):
def DFS(queens, xy_dif, xy_sum):
p = len(queens)
if p==n:
result.append(queens)
return None
for q in range(n):
if q not in queens and p-q not in xy_dif and p+q not in xy_sum:
DFS(queens+[q], xy_dif+[p-q], xy_sum+[p+q])
result = []
DFS([],[],[])
return [ ["."*i + "Q" + "."*(n-i-1) for i in sol] for sol in result]