[leetcode 19] N-Queens
2014-08-05 15:48
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题目:
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.."] ]
思路:
1.皇后问题,用dfs解决;2.可用一个int数组作为记录,num[i]=k表示第i行的棋子放在第k列;
3.dfs依次对每行(0至n-1)进行考虑,每行对每列(0至n-1)进行考虑,对每种可能性进行判断,即用一个bool型的可能性进行判断,如判断结果为真,即表示当前情况下第cur行的棋子可放在第i列,之后考虑cur+1行,对该行再对每列(0至n-1)进行考虑,如果在某种情况下cur==n,则表示该种情况可行。
4.check函数,分为三种情况(同列,正斜线,反斜线)。
代码:
class Solution{ public: vector<vector<string> > sloveNQueens(int n) { vector<vector<string> > result; vector<int> num(n,-1); dfs(result,num,0); return result; } private: void dfs(vector<vector<string> > &result,vector<int> &num,int cur) { //num:num[i]表示第i行放在第num[i]列 //cur:cur表示所考虑的当前行 if(cur==num.size()) { vector<string> tmp; for(int i=0;i!=num.size();++i) { string str(num.size(),'.'); str[num[i]]='Q'; tmp.push_back(str); } result.push_back(tmp); return; } //i表示第cur行可放的列数 //对于每个cur的值,要考虑所有的列数(i)的可能 for(int i=0;i!=num.size();++i) { if(check(num,cur,i)) { num[cur]=i; dfs(result,num,cur+1); } } } bool check(vector<int> num,int cur,int val) { //cur现在正考虑的行 //val现在欲在cur行存放的列 for(int i=0;i!=cur;++i) { if(num[i]==val) return false; if(i+num[i]==cur+val) return false; if(i-num[i]==cur-val) return false; } return true; } };
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