PattySung

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. ->有n個人，可能有一個法官存在，所有的人都會信任他，但法官不會信任任何人；也可能沒有法官存在。

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>

建立陣列，把1~n的index放進去，所以要建立new int [n+1]
由於[a,b]，前者信任後者，所以我們可以計算t[1]後者出現的次數，每計算到一次++；而再前者t[0]出現的話，就–
迴圈：如果count[i]==n-1，代表每個人都信任他，所以他是法官


class Solution {
    public int findJudge(int n, int[][] trust) {
        int [] count = new int [n+1];
        for(int[] t : trust){
            count[t[0]]--; //trust別人的人--
            count[t[1]]++; //被trust的人++
        }
        for(int i=1 ; i<=n ; i++){
            if(count[i]== n-1) return i;
        }
        return -1;
        
    }
}

Time: O(N) Space: O(N)

Leetcode-997. Find the Town Judge

<解題>