51. N-Queens (Hard)

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: Backtracking

经典的N皇后问题，基本所有的算法书中都会包含的问题，经典解法为回溯递归，一层一层的向下扫描，需要用到一个pos数组，其中pos[i]表示第i行皇后的位置，初始化为-1，然后从第0开始递归，每一行都一次遍历各列，判断如果在该位置放置皇后会不会有冲突，以此类推，当到最后一行的皇后放好后，一种解法就生成了，将其存入结果res中，然后再还会继续完成搜索所有的情况，代码如下：

version 1: 3ms

class Solution {
    void helper(vector<vector<string>>& res, int n, int r, vector<int>& pos) {
        if (r == n) {
            vector<string> out(n, string(n,'.'));
            for (int i = 0; i < n; ++i) out[i][pos[i]] = 'Q';
            res.push_back(out);
            return;
        }

        for (int col = 0; col < n; ++col) {
            int row = 0;
            for (; row < r; ++row) {
                if (pos[row] == col || abs(col-pos[row]) == abs(r-row)) break;
            }
            if (row == r) {
                pos[r] = col;
                helper(res, n, r+1, pos);
                pos[r] = -1;
            }
        }
    }
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<vector<string>> res;
        vector<int> pos(n,-1);
        helper(res, n, 0, pos);
        return res;
    }
};