leetcode--N-Queens

1.题目描述

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,figure below There exist two distinct solutions to the 4-queens puzzle: [ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ]

 

2.解法分析

这个是DFS的典型应用，回溯法来解这个题的代码如下：

class Solution { public: vector<vector<string> > solveNQueens(int n) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<vector<int> >result; vector<int> path; mySolveNQueens(result,path,n); //构建输出的每一项的模板，节省时间 string line(n,'.'); vector<string>square(n,line); vector<vector<string> >string_result; vector<vector<int> >::iterator iter; for(iter=result.begin();iter!=result.end();++iter) { vector<string> temp=square; for(int i=0;i<n;++i) { temp[i][(*iter)[i]]='Q'; } string_result.push_back(temp); } return string_result; } //深度搜索，不断回溯 void mySolveNQueens(vector<vector<int> > &result,vector<int> &path,int n) { //在当前层遍历 for(int i=0;i<n;++i) { //如果当前层的某个位置i和之前的摆放不冲突，则加入该位置 if(!isConflict(path,i)) { path.push_back(i); if(path.size()==n)result.push_back(path);//找到一个解，继续 else mySolveNQueens(result,path,n);//还没到最后一层，继续 path.pop_back();//考虑该层的下一个位置，当然，得先把这个位置抹掉 } } } bool isConflict(vector<int> &path,int loc) { for(int i=0;i<path.size();++i) { //在同一列或者同一斜线上为冲突 if(path[i]==loc||((path[i]-loc)==(path.size()-i))||(path[i]-loc)==(i-path.size()))return true; } return false; } };