(图论基础题) leetcode 997. Find the Town Judge

In a town, there are N people labelled from 1 to N.  There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

The town judge trusts nobody.Everybody (except for the town judge) trusts the town judge.There is exactly one person that satisfies properties 1 and 2.

You are given trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.

If the town judge exists and can be identified, return the label of the town judge.  Otherwise, return -1.

 

Example 1:

Input: N = 2, trust = [[1,2]] Output: 2

Example 2:

Input: N = 3, trust = [[1,3],[2,3]] Output: 3

Example 3:

Input: N = 3, trust = [[1,3],[2,3],[3,1]] Output: -1

Example 4:

Input: N = 3, trust = [[1,2],[2,3]] Output: -1

Example 5:

Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]] Output: 3

 

Note:

1 <= N <= 1000trust.length <= 10000trust[i] are all differenttrust[i][0] != trust[i][1]1 <= trust[i][0], trust[i][1] <= N

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PS：记得几个月前在leetcode周赛上做这个题时，当时没有好好学图论，觉得这个题怎么这么难，还是但是标签是easy。。。最后用了一个很麻烦的方法写了。现在，学了图论后，发现这个题是真的很easy。。。看来，我还需要更加努力了

这个题是图论题，就是找入度为N-1和出度为0的点，trust他人就是出度，被trust就是入度。所以，最后这个题可以化为入度和出度差值为N-1的人。没找到就返回-1。

C++代码：

class Solution { public: int findJudge(int N, vector<vector<int>>& trust) { vector<int> vec(N+1,0); for(auto t : trust){ vec[t[0]]--; vec[t[1]]++; } for(int i = 1; i <= N; i++){ if(vec[i] == N-1) return i; } return -1; } };

 

另外附上几个月前写的代码：

class Solution { public: int findJudge(int N, vector<vector<int>>& trust) { set<int> s1,s2; if(trust.size() == 0) return 1; int num; vector<vector<int> > ans; for(int i = 0; i < trust.size(); i++){ s1.insert(trust[i][0]); s2.insert(trust[i][1]); } for(set<int>::iterator it = s2.begin(); it!=s2.end(); it++){ if(s1.count(*it)) s2.erase(*it); } if(s2.size() != 1) return -1; else{ for(set<int>::iterator it = s2.begin(); it!=s2.end(); it++){ num = *it; } for(int i = 0;i < trust.size(); i++){ if(trust[i][1] == num){ ans.push_back(trust[i]); } } if(ans.size() == N - 1) return ans[0][1]; else return -1; } } };

 

转载于:https://www.cnblogs.com/Weixu-Liu/p/10883214.html