其他分享
首页 > 其他分享> > leetcode【51】N-Queens

leetcode【51】N-Queens

作者:互联网

问题描述:

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.

Example:

Input: 4
Output: [
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

源码:

八皇后问题。很典型的回溯法,没啥好说的。但是有两点要注意:

1.尽量弄全局变量

2.参数传递尽量中指针(即&)!参数传递尽量中指针(即&)!参数传递尽量中指针(即&)!

给大家看一下不用指针和用指针的差距

class Solution {
public:
    vector<vector<string>> result;
    vector<vector<string>> solveNQueens(int n) {
        vector<int> visit(n, -1);
        
        solve(0, visit);
        
        return result;
    }
    
    void solve(int now, vector<int> &visit){
        int n=visit.size();
        if(now==n){
            vector<string> tmp(n, string(n,'.'));
            for(int i = 0; i < n; i++)
                tmp[i][visit[i]] = 'Q';
            result.push_back(tmp);
            return;
        }
        for(int col=0; col<n; col++){
            if(isvalid(visit, col, now)){
                visit[now] = col;
                solve(now+1, visit);
                visit[now] = -1;
            }
        }
    }
    
    bool isvalid(vector<int> &visit, int col, int now){
        for(int i=0; i<now; i++){
            if(visit[i] == col ||  abs(now-i)==abs(col-visit[i]))
                return false;
        }
        return true;
    }
};

 

标签:...,..,int,51,visit,vector,Queens,leetcode,queens
来源: https://blog.csdn.net/fanyuwgy/article/details/104857228