#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:

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

代码：

class Solution {

public:

    vector<int> v[1010];

    int vis[1010];

    int findJudge(int N, vector<vector<int>>& 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;

    }

};

巴特西

#Leetcode# 997. Find the Town Judge

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