10/27 每日一題 找出法官, 法官不信任人 , 除了法官外都人民都信任法官

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:

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 an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi. If a trust relationship does not exist in trust array, then such a trust relationship does not exist.

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

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

 

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

 

class Solution:

    def findJudge(self, n: int, trust: List[List[int]]) -> int:

        k = len(trust)

        if k<1 :

            if n==1:

                return 1

            else:

                return -1

       

#找出得票的人中 沒有出現在投票人清單

        tick = {}

       

        tick_person=[]

        for i in trust:

            tick_person.append(i[0])

            tick[i[1]]=tick.get(i[1],0)+1

        for K,v in tick.items():

            print(f'k={K}, n={n},v={v}')

            if v==n-1 and K not in tick_person:

                return K

        return -1    

           -------------------------------------------

class Solution:
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        
        trustMap = [0 for _ in range(n+1)]
        trustedMap = [0 for _ in range(n+1)]

        for a,b in trust:
            trustMap[b] +=1
            trustedMap[a] +=1

        for i in range(1,n+1):
            if trustMap[i] == n -1 and trustedMap[i] == 0:
                return i

return -1