LeetCode 51.N-Queens

最新推荐文章于 2022-06-06 22:03:40 发布

原创最新推荐文章于 2022-06-06 22:03:40 发布 · 396 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#backtracking

LeetCode解题报告专栏收录该内容

52 篇文章

订阅专栏

本文详细介绍了使用递归深度优先搜索算法解决N皇后问题的方法，包括核心逻辑实现和代码解析，提供了两种有效解决方案。通过实例演示了如何在N×N棋盘上放置N个皇后，确保任意两个皇后之间没有攻击关系。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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 all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

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

分析与解答：递归的DFS，不算难，注意判断是否合理的条件。

class Solution{
    public: 
        vector<vector<string> > result;
        vector<vector<string> > solveNQueens(int n){
            vector<string> path(n,string(n,'.'));
            dfs(0,0,n,path);
            return result;
        }
        void dfs(int row,int col,int n,vector<string> &path){
            if(row == n){
                result.push_back(path);
                return;
            }
            for(int j = 0;j != n;++j){
                path[row][j] = 'Q';
                if(isSafe(path,n,row,j)){
                    dfs(row+1,0,n,path);
                }
                path[row][j] = '.';
            }
            return;
        }

        bool isSafe(vector<string> &path,int n,int row,int col){
            for(int i = 0;i != row;++i){
                if(path[i][col] == 'Q'){
                    return false;
                }
            }
            int i = row - 1, j = col - 1, k = col + 1;
            while((i < row&&i >=0) &&(j >= 0 ||k < n)){
                if(path[i][j--] == 'Q'|| path[i][k++] == 'Q'){
                    return false;
                }
                i--;
            }
            return true;
        }
};