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.."]
]

皇后不能互相攻击，满足同一行，同一列，对角，不能同时存在两个Queen.

我利用了在解决Permutation时的思路，采用swap选择每行的列。这样，可以简化isvalid的函数，它只需要验证是否对角上已经存在一个Queen就可以了。

此算法在leetcode上，实际执行时为9ms。

class Solution {
public:
    vector<vector<string> > solveNQueens(int n) {
        vector<vector<string> > ans;
        vector<int> solution;
        for (int i=0; i<n; i++)
                solution.push_back(i);

        helper(ans, solution, 0);
        return ans;
    }

    void helper(vector<vector<string> > &ans, vector<int> &solution, int start) {
        if (start == solution.size()) {
                ans.push_back(vector<string>());
                for (int i=0; i<solution.size(); i++) {
                        ans.back().push_back(string(solution.size(), '.'));
                        ans.back().back()[solution[i]] = 'Q';
                }
                return;
        }

        for (int i=start; i<solution.size(); i++) {
                swap(solution[start], solution[i]);
                if (isValid(solution, start)) {
                        helper(ans, solution, start+1);
                }
                swap(solution[start], solution[i]);
        }


    }

    bool isValid(vector<int> &solution, int index) {
        for (int row=index-1; row>=0; --row) {
                if (index-row == abs(solution[index] - solution[row]))
                        return false;
        }

        return true;
    }
};

同时我也在leetcode上分享了上面这段代码：N Queens

N-Queens -- leetcode