Leetcode算法题目：N-Queens

题目：

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:

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分析：

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。

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代码：

class Solution {public:    vector<vector<string> > solveNQueens(int n) {        vector<vector<string> > res;        vector<int> pos(n, -1);        solveNQueensDFS(pos, 0, res);        return res;    }    void solveNQueensDFS(vector<int> &pos, int row, vector<vector<string> > &res) {        int n = pos.size();        if (row == n) {            vector<string> out(n, string(n, '.'));            for (int i = 0; i < n; ++i) {                out[i][pos[i]] = 'Q';            }            res.push_back(out);        } else {            for (int col = 0; col < n; ++col) {                if (isValid(pos, row ,col)) {                    pos[row] = col;                    solveNQueensDFS(pos, row + 1, res);                    pos[row] = -1;                }            }        }    }    bool isValid(vector<int> &pos, int row, int col) {        for (int i = 0; i < row; ++i) {            if (col == pos[i] || abs(row - i) == abs(col - pos[i])) {                return false;            }        }        return true;    }};