[Leetcode 997] Find the Town Judge

原题说明

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:

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 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. 1 <= N <= 1000
2. trust.length <= 10000
3. trust[i] are all different
4. trust[i][0] != trust[i][1]
5. 1 <= trust[i][0], trust[i][1] <= N

解题思路

这是一道单向图的题，实际上我们要寻找图中入度为N-1,出度为0的点。
因为每个点的入度最大即为N-1（即村上所有其他人都trust他）,所以入度 - 出度 = N - 1是入度为N-1,出度为0的充分必要条件，所以本题用一个counts数组统计每个点入度和出度之差，遍历trust来更新counts数组，之后再遍历counts数组寻找值为N-1的点，返回坐标即可，如果在counts中不存在值为N-1的点，说明不存在judge，返回-1。

示例代码 (cpp)

class Solution {
public:
    int findJudge(int N, vector<vector<int>>& trust) {
        vector<int> counts(N + 1, 0);
        for (const auto& relation : trust) {
            counts[relation[0]]--;
            counts[relation[1]]++;
        }
        for (int i = 1; i <= N; ++i) {
            if (counts[i] == N - 1) {
                return i;
            }
        }
        return -1;
    }
};

复杂度分析

时间复杂度: O(T + N), T为trust数组的长度
空间复杂度: O(N)

归纳总结

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