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;
    }
};