# (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:

1. The town judge trusts nobody.
2. Everybody (except for the town judge) trusts the town judge.
3. 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. 1 <= N <= 1000
2. trust.length <= 10000
3. trust[i] are all different
4. trust[i][0] != trust[i][1]
5. 1 <= trust[i][0], trust[i][1] <= N

==================================================================

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

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;
}
};

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;
}
}
};

posted @ 2019-05-17 19:19  PJCK  阅读(15)  评论(0编辑  收藏