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 resfillRes(indices,n);return;} // loop each possible one if valid nextfor (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;};