【数据结构基础_图】Leetcode 997.找到小镇的法官

原题链接：Leetcode 997. Find the Town Judge

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.

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
1
2

Example 2:

Input: n = 3, trust = [[1,3],[2,3]]
Output: 3
1
2

Example 3:

Input: n = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1
1
2

Constraints:

• 1 <= n <= 1000
• 0 <= trust.length <= 10⁴
• trust[i].length == 2
• All the pairs of trust are unique.
• ai != bi
• 1 <= ai, bi <= n

---

方法一：统计入度出度

思路：

这里的信任就是有向边
代表a信任b
法官不信任任何人，所有人都信任法官，则法官的出度为0，入度为n-1。找到这样的结点并返回即可

C++代码：

class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
        // 记录入度和出度
        vector<int> in(n + 1);
        vector<int> out(n + 1);

        // 读取每条边 记录端点的入度和出度
        for (auto& edge : trust) {
            int x = edge[0], y = edge[1];
            in[y]++;
            out[x]++;
        }

        for (int i = 1; i <= n; ++i) {
            if (in[i] == n - 1 && out[i] == 0) {
                return i;
            }
        }
        return -1;
    }
};

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

复杂度分析：

• 时间复杂度：O(n)，需要遍历每个结点
• 空间复杂度：O(n)，两个数组记录入度和出度