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.."]]

一道深度搜索题目，用三个数组表示主对角线，副对角线，列，有没有queen占据，用此信息剪枝

Have you been asked this question in an interview? 

class Solution {public:    void dfs(vector<vector<string> >&result,vector<int> path,vector<int> diag1,vector<int> diag2,vector<int> col,int n){        if(path.size()==n){            vector<string> ans;            for(int i=0;i<n;i++){                string temp(n,'.');                temp[path[i]]='Q';                ans.push_back(temp);            }            result.push_back(ans);            return;        }        for(int i=0;i<n;i++){            if(col[i]==0&&diag1[path.size()+i]==0&&diag2[path.size()+n-1-i]==0){                col[i]=diag1[path.size()+i]=diag2[path.size()+n-1-i]=1;                path.push_back(i);                dfs(result,path,diag1,diag2,col,n);                path.pop_back();                col[i]=diag1[path.size()+i]=diag2[path.size()+n-1-i]=0;            }        }    }    vector<vector<string> > solveNQueens(int n) {        vector<vector<string> > result;        vector<int> path;        vector<int> diag1(2*n-1,0);        vector<int> diag2(2*n-1,0);        vector<int> col(n,0);        dfs(result,path,diag1,diag2,col,n);        return result;    }};