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.."]
]

没做出来，看了yu的答案写的。不过确实是backtracking的思想： 如果到了符合条件的记录，如果不到，那么尝试每一个可能的子集，如果当前没有问题，那么继续下一个判断

class Solution {
public:
	void fillRes(vector<int>& indices,int n)
	{
		vector<string> r;
		for(int i=0; i<n; i++){
			string s(n,'.');
			s[indices[i]]='Q';
			r.push_back(s);
		}
		res.push_back(r);
	}
	bool isValid(vector<int>& indices, int c)
	{

<span style="white-space:pre">		</span>//queen cannot sit on the same col or diagonal or row

		for(int i=0; i<c;i++){
			if(indices[i]==indices[c]|| (abs(indices[i]-indices[c])==c-i))
				return false;
		}
		return true;
	}
	void nqueens(vector<int>& indices, int c, int n)
	{
		if (c==n){

<span style="white-space:pre">	</span>                //the last one, push to res
			fillRes(indices,n);
			return;
		} // loop each possible one if valid next
		for (int i=0; i<n; i++){
			indices[c]=i;
			if(isValid(indices,c))
				nqueens(indices,c+1,n);
		}
	}
    vector<vector<string> > solveNQueens(int n) {
        res.clear();
        vector<int> indices(n,-1);
        nqueens(indices,0,n);
        return res;
    }
private:
    vector<vector<string> > res;
};