![Leetcode 51. N-Queens [Python],第1张 Leetcode 51. N-Queens [Python],第1张](/aiimages/Leetcode+51.+N-Queens+%5BPython%5D.png)
回溯法,当遍历到最后一行,此时的组合就可以被记录了。而如果不到的话,还需要为当前的行的queen选择合适的列。注意设置对脚线数组记录选定当前行当前列后,向左上的对角线和向右上的对角线,判断是否合法。其位置的计算是当前行数+选定列数,和总行数N-当前行数+选定列数。
class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
col = [False]*n
udg = [False]*(2*n)
dg = [False]*(2*n)
res = []
grid = [['.' for _ in range(n)] for _ in range(n)]
def dfs(currow):
if currow == n:
tempres = []
for i in range(n):
line = ''
for j in range(n):
line += grid[i][j]
tempres.append(line)
res.append(tempres)
return
else:
for i in range(n):
if not col[i] and not udg[currow + i] and not dg[n - currow + i]:
col[i] = udg[currow + i] = dg[n - currow + i] = True
grid[currow][i] = 'Q'
dfs(currow + 1)
grid[currow][i] = '.'
col[i] = udg[currow + i] = dg[n - currow + i] = False
dfs(0)
return res
欢迎分享,转载请注明来源:内存溢出
微信扫一扫
支付宝扫一扫
评论列表(0条)