★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號:山青詠芝(shanqingyongzhi)
➤博客園地址:山青詠芝(https://www.cnblogs.com/strengthen/)
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:http://www.javashuo.com/article/p-ebbvzqqw-md.html
➤若是連接不是山青詠芝的博客園地址,則多是爬取做者的文章。
➤原文已修改更新!強烈建議點擊原文地址閱讀!支持做者!支持原創!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html
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.git
If the town judge exists, then:github
You are given trust
, an array of pairs trust[i] = [a, b]
representing that the person labelled a
trusts the person labelled b
.數組
If the town judge exists and can be identified, return the label of the town judge. Otherwise, return -1
.微信
Example 1:ide
Input: N = 2, trust = [[1,2]] Output: 2
Example 2:spa
Input: N = 3, trust = [[1,3],[2,3]] Output: 3
Example 3:code
Input: N = 3, trust = [[1,3],[2,3],[3,1]] Output: -1
Example 4:htm
Input: N = 3, trust = [[1,2],[2,3]] Output: -1
Example 5:blog
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 differenttrust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N
在一個小鎮裏,按從 1
到 N
標記了 N
我的。傳言稱,這些人中有一個是小鎮上的祕密法官。
若是小鎮的法官真的存在,那麼:
給定數組 trust
,該數組由信任對 trust[i] = [a, b]
組成,表示標記爲 a
的人信任標記爲 b
的人。
若是小鎮存在祕密法官而且能夠肯定他的身份,請返回該法官的標記。不然,返回 -1
。
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]
是徹底不一樣的trust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N
776ms
1 class Solution { 2 func findJudge(_ N: Int, _ trust: [[Int]]) -> Int { 3 if( N == 1) { 4 return N 5 } 6 if(trust.count == 0 || trust[0].count==0){ 7 return -1; 8 } 9 var vote = [Int](repeating: 0, count: N) 10 for i in 0..<trust.count { 11 var trustee = trust[i][0] //someone is a trustee 12 var trusted = trust[i][1] // Trusted by someone 13 vote[trustee - 1] = vote[trustee - 1] - 1 14 vote[trusted - 1] = vote[trusted - 1] + 1 15 } 16 for i in 0..<N{ 17 if(vote[i] == N-1) { 18 return i + 1 19 } 20 } 21 return -1 22 } 23 }
784ms
1 class Solution { 2 func findJudge(_ N: Int, _ trust: [[Int]]) -> Int { 3 var arr = Array(repeating: 0, count: N + 1) 4 for i in 0..<trust.count { 5 arr[trust[i][1]] += 1 6 arr[trust[i][0]] -= 1 7 } 8 for i in 1..<arr.count { 9 if arr[i] == N - 1 {return i} 10 } 11 return -1 12 } 13 }
788ms
1 class Solution { 2 func findJudge(_ N: Int, _ trust: [[Int]]) -> Int { 3 var res = [Int].init(repeating: 0, count: N) 4 var trusts = [Bool].init(repeating: false, count: N) 5 for pair in trust { 6 res[pair[1] - 1] += 1 7 trusts[pair[0] - 1] = true 8 } 9 10 for i in 0..<N { 11 if res[i] == N - 1 && trusts[i] == false { 12 return i + 1 13 } 14 } 15 return -1 16 } 17 }
800ms
1 class Solution { 2 func findJudge(_ N: Int, _ trust: [[Int]]) -> Int { 3 var candDict: [Int: Int] = [:] 4 for index in 1...N+1{ 5 candDict[index] = 0 6 } 7 for tru in trust{ 8 var t = tru[0] 9 var r = tru[1] 10 if let val = candDict[t]{ 11 candDict[t] = nil 12 } 13 if let val = candDict[r]{ 14 candDict[r]! += 1 15 } 16 } 17 for (key,val) in candDict{ 18 if val == N-1{ 19 return key 20 } 21 } 22 return -1 23 } 24 }
816ms
1 class Solution { 2 func findJudge(_ N: Int, _ trust: [[Int]]) -> Int { 3 var inAndOut = [[Int]](repeating: [Int](repeating: 0, count: 2),count: N + 1) 4 for t in trust { 5 inAndOut[t[0]][0] += 1 6 inAndOut[t[1]][1] += 1 7 } 8 9 for i in 1..<inAndOut.count { 10 let item = inAndOut[i] 11 if item[1] == N - 1 && item[0] == 0 { 12 return i 13 } 14 } 15 return -1 16 } 17 }