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.."] ]
class Solution { private: std::vector<std::vector<std::string> > res; public: std::vector<std::vector<std::string> > solveNQueens(int n) { std::vector<std::string>cur(n, std::string(n,'.')); dfs(cur, 0); return res; } void dfs(std::vector<std::string> &cur, int row) { if(row == cur.size()) { res.push_back(cur); return; } for(int col = 0; col < cur.size(); col++) if(isValid(cur, row, col)) { cur[row][col] = 'Q'; dfs(cur, row+1); cur[row][col] = '.'; } } //推断cur[row][col]位置放置皇后。是否合法。 bool isValid(std::vector<std::string> &cur, int row, int col) { //列 for(int i = 0; i < row; i++) if(cur[i][col] == 'Q')return false; //右对角线 for(int i = row-1, j=col-1; i >= 0 && j >= 0; i--,j--) if(cur[i][j] == 'Q')return false; //左对角线 for(int i = row-1, j=col+1; i >= 0 && j < cur.size(); i--,j++) if(cur[i][j] == 'Q')return false; return true; } };
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