997. Find the Town Judge

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本文介绍了一种算法，用于在一个标记从1到N的人群中找出可能存在的小镇法官。法官不信任任何人，但除了他自己外，每个人都信任他。通过解析信任关系数组，文章详细解释了如何确定满足条件的法官是否存在及其身份。

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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

题解：别看题目有点长，但其实都是唬人的。把trust数组遍历一遍，建立一个矩阵，但是不同长，然后在遍历一遍矩阵，如果他的flag为true，就代表他有信任的人，就说明他不是town judge，如果有且仅有一个person，信任他的人数是N-1，并且他的flag不为true，则返回1，否则返回-1

/**
 * @param {number} N
 * @param {number[][]} trust
 * @return {number}
 */
var findJudge = function(N, trust) {
   let flag = new Array(N)
   let tr = new Array(N)
   let len = trust.length
   for(let i = 1; i <= N; i ++) {
       tr[i] = []
   }
   for(let i = 0; i < len; i ++) {
       flag[trust[i][0]] = true
       if(!flag[trust[i][1]]) {
           tr[trust[i][1]].push(trust[i][0])
       }
   }
    let ans = -1
    for(let i = 1; i <= N; i ++) {
        if(!flag[i] && tr[i].length === N - 1) {
            if(ans !== -1) return -1
            else ans = i
        } 
    }
    return ans
};