997 Find the Town Judge 找到小鎮(zhèn)的法官
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
題目描述:
在一個(gè)小鎮(zhèn)里糊啡，按從 1 到 N 標(biāo)記了 N 個(gè)人。傳言稱炫刷，這些人中有一個(gè)是小鎮(zhèn)上的秘密法官。
如果小鎮(zhèn)的法官真的存在，那么：
小鎮(zhèn)的法官不相信任何人雪侥。
每個(gè)人（除了小鎮(zhèn)法官外）都信任小鎮(zhèn)的法官彰亥。
只有一個(gè)人同時(shí)滿足屬性 1 和屬性 2 葱色。
給定數(shù)組 trust卸伞，該數(shù)組由信任對 trust[i] = [a, b] 組成抹镊，表示標(biāo)記為 a 的人信任標(biāo)記為 b 的人。
如果小鎮(zhèn)存在秘密法官并且可以確定他的身份瞪慧，請返回該法官的標(biāo)記。否則部念，返回 -1弃酌。
示例 :
示例 1：
輸入：N = 2, trust = [[1,2]]
輸出：2
示例 2：
輸入：N = 3, trust = [[1,3],[2,3]]
輸出：3
示例 3：
輸入：N = 3, trust = [[1,3],[2,3],[3,1]]
輸出：-1
示例 4：
輸入：N = 3, trust = [[1,2],[2,3]]
輸出：-1
示例 5：
輸入：N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
輸出：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:
遍歷 trust數(shù)組, 記錄出度和入度的差, 返回出度和入度差為 N - 1的人
時(shí)間復(fù)雜度O(n), 空間復(fù)雜度O(n)
代碼:
C++:
class Solution
{
public:
int findJudge(int N, vector<vector<int>>& trust)
{
vector<int> count(N, 0);
for (const 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
