52. N-Queens II-优快云博客

本文详细探讨了N皇后问题，即如何在N×N的棋盘上放置N个皇后，使得任意两个皇后不能互相攻击。通过递归回溯算法，实现了N皇后问题的有效求解，并提供了具体的C++代码实现。

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.."]
]

AC code:

class Solution {
public:
    int totalNQueens(int n) {
        int res = 0;
        vector<vector<string>> v;
        vector<string> nqueens(n, string(n, '.'));
        solve(v, nqueens, n, 0, res);
        return res;
    }
     
    void solve(vector<vector<string>>& v, vector<string>& nqueens, int n, int row, int& res) {
        if (row == n) {
            res++;
            return;
        }
        for (int col = 0; col < n; ++col) {
            if (judge(nqueens, n, row, col)) {
                nqueens[row][col] = 'Q';
                solve(v, nqueens, n, row+1, res);
                nqueens[row][col] = '.';
            }
        }
    }
 
    bool judge(vector<string> nqueens, int n, int x, int y) {
        // up and down
        for (int i = 0; i < n && i != x; ++i) {
            if (nqueens[i][y] == 'Q') return false;
        }
        // right and left
        for (int i = 0; i < n && i != y; ++i) {
            if (nqueens[x][i] == 'Q') return false;
        }
        // left up
        for (int i = x-1, j = y-1; i >= 0 && j >= 0; --i, --j) {
            if (nqueens[i][j] == 'Q') return false;
        }
        // right up
        for (int i = x-1, j = y+1; i >= 0 && j < n; --i, ++j) {
            if (nqueens[i][j] == 'Q') return false;
        }
        return true;
    }
};

Runtime: 16 ms, faster than 7.86% of C++ online submissions for N-Queens II.

转载于:https://www.cnblogs.com/ruruozhenhao/p/9813638.html