N queen

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.."]
]

1. back tracking method

  class Solution {
public:
    vector<vector<string> > solveNQueens(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<string> > results;
        if (n<=0)
            return results;
            
        vector<string> cur;
        solve(results,cur,n,0);
    }
    
    void solve(vector<vector<string> >& results,vector<string> cur,int n,int index){
        if (index == n){
            results.push_back(cur);
            return;
        }
        
        bool * table= new bool[n];
        for (int i = 0; i<n;i++)
            table[i]= true;
        
        for(int m = 0; m<index; m++)
            for (int i = 0; i<n ; i++){
                if (cur[m][i] == 'Q'){
                    table[i] = false;
                    if (i - (index-m) >=0)
                        table[i - (index-m)] =false;
                    if (i + (index-m) < n)
                        table[i + (index-m)] =false;
                }
            }
        string temp;   
        for (int i = 0; i<n; i++){
            temp += '.';
        }
        
        for (int i = 0; i<n ;i++){
            if (table[i] == true){
                temp[i] = 'Q';
                cur.push_back(temp);
                solve(results,cur,n,index+1);
                cur.pop_back();
                temp[i] = '.';
            }
        }
        delete [] table;
    }
    
};

2. 可以用排列组合(permutation and combination)，全排列所有的情况，然后判断 i-j == Cur[i]-Cur[j] or j-i == Cur[i]-Cur[j]