997. Find the Town Judge

⭐

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:

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 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;
  }
};