Description:
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:
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
题目描述:
在一个小镇里，按从 1 到 N 标记了 N 个人。传言称，这些人中有一个是小镇上的秘密法官。
如果小镇的法官真的存在，那么：
小镇的法官不相信任何人。
每个人（除了小镇法官外）都信任小镇的法官。
只有一个人同时满足属性 1 和属性 2 。
给定数组 trust，该数组由信任对 trust[i] = [a, b] 组成，表示标记为 a 的人信任标记为 b 的人。
如果小镇存在秘密法官并且可以确定他的身份，请返回该法官的标记。否则，返回 -1。
示例 :
示例 1：
输入：N = 2, trust = [[1,2]]
输出：2
示例 2：
输入：N = 3, trust = [[1,3],[2,3]]
输出：3
示例 3：
输入：N = 3, trust = [[1,3],[2,3],[3,1]]
输出：-1
示例 4：
输入：N = 3, trust = [[1,2],[2,3]]
输出：-1
示例 5：
输入：N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
输出：3
提示：
1 <= N <= 1000
trust.length <= 10000
trust[i] 是完全不同的
trust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N
思路:
遍历 trust数组, 记录出度和入度的差, 返回出度和入度差为 N - 1的人
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
int findJudge(int N, vector>& trust)
{
int count[N] = {0};
for (auto person : trust)
{
count[person[0] - 1]--;
count[person[1] - 1]++;
}
for (int i = 0; i < N; i++) if (count[i] == N - 1) return i + 1;
return -1;
}
};
Java:
class Solution {
public int findJudge(int N, int[][] trust) {
int count[] = new int[N];
for (int[] person : trust) {
count[person[0] - 1]--;
count[person[1] - 1]++;
}
for (int i = 0; i < N; i++) if (count[i] == N - 1) return i + 1;
return -1;
}
}
Python:
class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
count = [0] * N
for person in trust:
count[person[0] - 1] -= 1
count[person[1] - 1] += 1
for i, person in enumerate(count):
if person == N - 1:
return i + 1
return -1