N-Queens -- leetcode
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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
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