N-QueensII使用回溯很慢

n-皇后拼图是将n 个皇后放在nxn 棋盘上的问题，这样就不会有两个皇后互相攻击。

给定一个整数 n，返回 n-queens 谜题的不同解的数量。

https://leetcode.com/problems/n-queens-ii/

我的解决方案：

class Solution:
    def totalNQueens(self, n: int) -> int:
        def genRestricted(restricted, r, c):
            restricted = set(restricted)
            for row in range(n): restricted.add((row, c))
            for col in range(n): restricted.add((r, col))
            movements = [[-1, -1], [-1, 1], [1, -1], [1, 1]]
            for movement in movements:
                row, col = r, c
                while 0 <= row < n and 0 <= col < n:
                    restricted.add((row, col))
                    row += movement[0]
                    col += movement[1]
            return restricted

        def gen(row, col, curCount, restricted):
            count, total_count = curCount, 0

            for r in range(row, n):
                for c in range(col, n):
                    if (r, c) not in restricted:
                        count += 1
                        if count == n: total_count += 1
                        total_count += gen(row + 1, 0, count, genRestricted(restricted, r, c))
                        count -= 1

            return total_count

        return gen(0, 0, 0, set())

它在 n=8 时失败。我不知道为什么，以及如何减少迭代。看来我已经在做尽可能少的迭代了。

回答

def square_is_safe(file, rank, queens_placed):
    for queen_rank, queen_file in enumerate(queens_placed):
        if queen_file == file:                      # vertical attack
            return false
        if queen_file - file == queen_rank - rank:  # diagonal attack
            return false
        if queen_file - file == rank - queen_rank:  # anti-diagonal attack
            return false
    return true

def place_queen_at_rank(queens_placed, rank):
    if rank == n:
        total_count += 1
        return

    for file in range(0, n):
        if square_is_safe(file, rank, queens_placed):
            queens_placed.append(file)
            place_queen_at_rank(queens_placed, rank + 1)

    queens_placed.pop()