52. N-Queens II

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本文探讨了N皇后问题的第二种形式，即不关注具体的棋盘布局而只计算可能的解决方案数量。通过递归回溯的方法，实现了一个Python程序来解决这个问题，并给出了当N为4时的两种不同解法。

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

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 the number of distinct solutions to the n-queens puzzle.

Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

class Solution(object):
    def isValid(self, board, row, col):
        if board[row][col] == 'Q':
            return False
        n = len(board)
        for i in range(n):
            if board[i][col] == 'Q':
                return False
            if board[row][i] == 'Q':
                return False
        i = 1
        while row + i < n or row - i >= 0 or col + i < n or col - i >= 0:
            if row + i < n and col + i < n and board[row + i][col + i] == 'Q':
                return False
            if row + i < n and col - i >= 0 and board[row + i][col - i] == 'Q':
                return False
            if row - i >= 0 and col + i < n and board[row - i][col + i] == 'Q':
                return False
            if row - i >= 0 and col - i >= 0 and board[row - i][col - i] == 'Q':
                return False
            i += 1
        return True

    def solveNQueens(self, n):
        """
        :type n: int
        :rtype: List[List[str]]
        """
        self.res = 0
        board = [['.'] * n for _ in range(n)]
        self.solveBoard(board, 0)
        return self.res


    def solveBoard(self, board, col):
        n = len(board)
        if col == n:
            self.res += 1
            return
        for i in range(n):
            if self.isValid(board, i, col):
                board[i][col] = 'Q'
                self.solveBoard(board, col + 1)
                board[i][col] = '.'