#Leetcode# 997. Find the Town Judge

https://leetcode.com/problems/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

代码：

class Solution {
public:
    vector v[1010];
    int vis[1010];
    int findJudge(int N, vector>& trust) {
        int n = trust.size();
        
        for(int i = 0; i < n; i ++) {
            int a = trust[i][0], b = trust[i][1];
            v[b].push_back(a);
            vis[a] = 1;
        }
        
        int cnt = 0, temp = 0;
        for(int i = 1; i <= N; i ++) {
            if(vis[i] == 0 && v[i].size() == N - 1) {
                cnt ++;
                temp = i;
            }
        }
        if(cnt == 1) return temp;
        return -1;
    }
};

 

转载于:https://www.cnblogs.com/zlrrrr/p/10715985.html