997. Find the Town Judge

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# 題目敘述

In a town, there are n people labeled 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 an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi . If a trust relationship does not exist in trust array, then such a trust relationship does not exist.

Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.

# 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

# 解題思路

# Solution

class Solution {

    public int findJudge(int n, int[][] trust) {

        int[] townJudge = new int[n];

        int[] hastrust = new int[n];

        for (int i = 0; i < trust.length; i++) {

            hastrust[trust[i][0] - 1]++;

            townJudge[trust[i][1] - 1]++;

        }

        for (int i = 0; i < n; i++) {

            if (townJudge[i] == n - 1 && hastrust[i] == 0) {

                return i + 1;

            }

        }

        return -1;

    }

}

#include <vector>

#include <string.h>

using namespace std;

class Solution {

 public:

  int findJudge(int n, vector<vector<int>>& trust) {

    int townJudge[n];

    memset(townJudge, 0, n * sizeof(int));

    int hasTrust[n];

    memset(hasTrust, 0, n * sizeof(int));

    for (int i = 0; i < trust.size(); i++) {

      hasTrust[trust[i][0] - 1]++;

      townJudge[trust[i][1] - 1]++;

    }

    for (int i = 0; i < n; i++) {

      if (townJudge[i] == n - 1 && hasTrust[i] == 0) {

        return i + 1;

      }

    }

    return -1;

  }

};