N-Queens N皇后放置问题 回溯法

N-Queens

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.."]
]

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Backtracking

class Solution {
public:
    vector<vector<string> > res;
    
    vector<vector<string>> solveNQueens(int n) {
        
        int *a=new int
;
        memset(a,0,sizeof(int)*n);
        solve(a,n,0);
        return  res;
    }
    
    void solve(int *a, int n,int index)
    {
        for(int i=0;i<n;i++)//对每个index 寻找第i列放皇后
        {
            if(isValid(a,n,index,i))//index i 行 列
            {
                a[index]=i;
                if(index==n-1)
                {
                    print(a,n);
                    a[index]=0;
                    return ;
                }
                solve(a,n,index+1);
                a[index]=0;
            }
        }
    }
    
    void print(int *a, int n)
    {
        vector<string> path;
        for(int i=0;i<n;i++)
        {
            string s(n,'.');
            s[a[i]]='Q';
            path.push_back(s);
        }
        res.push_back(path);
    }
    
    bool isValid(int *a,int n,int x,int y)
    {
        int col;
        for(int i=0;i<x;i++)//行
        {
            col=a[i];//第i行，皇后在第col列
            if(y==col) //同列
                return false;
            if((col-y)==(i-x))// 对角线'\'
                return false;
            if((col-y)==(x-i))// 对角线'/'
                return false;
        }
        return true;
    }
};