leetcode-51-N皇后

题目描述：

 第一次提交：回溯：

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        def sub(string, p, c):
            new = [s for s in string]
            new[p] = c
            return ''.join(new)
        def check(i,j,board):
            for x in range(n):
                if board[x][j] == "Q":
                    return False
            for x in [-1,1]:
                for y in [-1,1]:
                    i1 = i + x
                    j1 = j + y
                    while 0<=i1<n and 0<=j1<n:
                        if board[i1][j1] == "Q":
                            return False
                        else:
                            i1 += x
                            j1 += y
            return True
        def backtrack(i,board):
            if i == n:
                res.append(board.copy())
                return
            for j in range(n):
                if check(i,j,board):
                    board[i] = sub(board[i],j,"Q")
                    backtrack(i+1,board)
                    board[i] = sub(board[i], j, ".")
            return False

        res = []
        board = ["." * n for _ in range(n)]
        backtrack(0, board)
        return res

优化：O(N!)

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        res = []
        s = "." * n
        def backtrack(i, tmp,col,z_diagonal,f_diagonal):
            if i == n:
                res.append(tmp)
                return 
            for j in range(n):
                if j not in col and i + j not in  z_diagonal and i - j not in f_diagonal:
                    backtrack(i+1,tmp + [s[:j] + "Q" + s[j+1:]], col | {j}, z_diagonal |{i + j} , f_diagonal |{i - j}  ) 
            
        backtrack(0,[],set(),set(),set())    
        return res

标签：set,return,backtrack,List,51,diagonal,皇后,leetcode,def
来源： https://www.cnblogs.com/oldby/p/11697714.html