Problem statement:

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.."]]

Solution:

As a classical, N-Queens typically is a DFS problem. The solution follows DFS template. 

According to the rules to place the queen, they can not in the same column, we do not need to keep a vector which records if a position is visited or not. 

class Solution {public:    vector
     
      
       > solveNQueens(int n) {        vector
       
        
         > solu_sets;        vector
         
           solu(n, string(n, '.'));        NQueens(solu_sets, solu, 0, n, n);        return solu_sets;    }private:    void NQueens(vector
          
           
            >& solu_sets, vector
            
             & solu, int row, int col, int n){ if(row == n){ solu_sets.push_back(solu); return; } for(int i = 0; i < col; i++){ if(isvalid(solu, row, i, n)){ solu[row][i] = 'Q'; NQueens(solu_sets, solu, row + 1, col, n); solu[row][i] = '.'; } } return; } bool isvalid(vector
             
              & solu, int row, int col, int n){ for(int i = row - 1, left = col - 1, right = col + 1; i >= 0; i--, left--, right++){ // vertically/diagonal/anti-diagonal conflict if(solu[i][col] == 'Q' || (left >= 0 && solu[i][left] == 'Q') || (right < n && solu[i][right] == 'Q')){ return false; } } return true; }};