[Leetcode]52. N-Queens II(C++)

题目描述

The n-queens puzzle is the problem of placing n queens on an n x 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.

例子

例子 1

见官网例子。

Constraints

1 <= n <= 9

解题思路

思路和 [Leetcode]51. N-Queens(C++) 基本一样，只是不需要存储具体的解，只需要用一个整型存储解的数量，在找到解时将其加一即可。代码如下：

#include <string>
#include <unordered_set>
#include <vector>

class Solution {
public:
    int totalNQueens(int n) {
        int result = 0;
        std::unordered_set<int> col_used;
        std::unordered_set<int> diag1;
        std::unordered_set<int> diag2;
        dfs(0, n, col_used, diag1, diag2, result);
        return result;
    }

private:
    void dfs(int row, int n, std::unordered_set<int>& col_used,
             std::unordered_set<int>& diag1, std::unordered_set<int>& diag2,
             int& result) {
        if (row == n) {
            result++;
            return;
        }

        std::string curr_row(n, '.');
        for (int col = 0; col < n; ++col) {
            if (col_used.count(col) || diag1.count(col - row) ||
                diag2.count(col + row))
                continue;
            curr_row[col] = 'Q';
            col_used.insert(col);
            diag1.insert(col - row);
            diag2.insert(col + row);
            dfs(row + 1, n, col_used, diag1, diag2, result);
            col_used.erase(col);
            diag1.erase(col - row);
            diag2.erase(col + row);
        }
    }
};

时间复杂度: O(n!)
空间复杂度: O(n)

GitHub 代码同步地址： 52.NQueensIi.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions