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.."] ]
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：
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; } };
本文转自博客园Grandyang的博客，原文链接：N皇后问题[LeetCode] N-Queens ，如需转载请自行联系原博主。
