[Swift]LeetCode274.H指数 | H-Index

发布时间：2020-12-14 05:06:56 所属栏目：百科来源：网络整理

导读：Given an array of citations (each citation is a non-negative integer) of a researcher,write a function to compute the researcher‘s h-index. According to the?definition of h-index on Wikipedia: "A scientist has index? h ?if? h ?of his/her?

Given an array of citations (each citation is a non-negative integer) of a researcher,write a function to compute the researcher‘s h-index.

According to the?definition of h-index on Wikipedia: "A scientist has index?h?if?h?of his/her?N?papers have?at least?h?citations each,and the other?N ? h?papers have?no more than?h?citations each."

Example:

Input: 
Output: 3 
Explanation: means the researcher has  papers in total and each of them had 
             received  citations respectively. 
?            Since the researcher has  papers with at least  citations each and the remaining 
?            two with no more than  citations each,her h-index is .citations = [3,6,1,5][3,5]53,53333

Note:?If there are several possible values for?h,the maximum one is taken as the h-index.

给定一位研究者论文被引用次数的数组（被引用次数是非负整数）。编写一个方法，计算出研究者的?h?指数。

h 指数的定义: “h 代表“高引用次数”（high citations），一名科研人员的 h 指数是指他（她）的（N 篇论文中）至多有 h 篇论文分别被引用了至少?h 次。（其余的?N - h?篇论文每篇被引用次数不多于?h?次。）”

示例:

输入: 
输出: 3 
解释: 给定数组表示研究者总共有  篇论文，每篇论文相应的被引用了  次。
?    由于研究者有 篇论文每篇至少被引用了  次，其余两篇论文每篇被引用不多于  次，所以她的 h 指数是 。citations = [3,53333

24ms

 1 class Solution {
 2     func hIndex(_ citations: [Int]) -> Int {
 3         let arr = citations.sorted { (num1,num2) -> Bool in
 4             return num1 < num2
 5         }
 6         for i in 0 ..< arr.count {
 7             if arr[i] >= arr.count - i {
 8                 return arr.count - i
 9             }
10         }
11         
12         return 0
13     }
14 }

28ms

 1 class Solution {
 2     func hIndex(_ citations: [Int]) -> Int {
 3         if citations.isEmpty {
 4             return 0
 5         }
 6         let n = citations.count
 7         var res = Array(repeating: 0,count: n + 1)
 8         
 9         for i in citations {
10             if i > n{
11                 res[n] += 1
12             }else {
13                 res[i] += 1
14             }
15         }
16         
17         for i in stride(from: n,to: 0,by: -1) {
18             if res[i] >= i {
19                 return i
20             }
21             res[i-1] += res[i]
22         }
23         
24         
25         return 0
26     }
27 }

28ms

 1 class Solution {
 2     func hIndex(_ citations: [Int]) -> Int {
 3         if citations.isEmpty {
 4             return 0
 5         }
 6         let n = citations.count
 7         var res = Array(repeating: 0,by: -1) {
18             if res[i] >= i {
19                 return i
20             }
21             res[i-1] += res[i]
22         }
23         
24         
25         return 0
26     }
27 }

36ms

 1 class Solution {
 2     func hIndex(_ citations: [Int]) -> Int {
 3         let n = citations.count
 4         if (n == 0) {return 0}
 5         var array = Array(repeating:0,count:n + 1)
 6         for i in 0..<n {
 7             if(citations[i] > n) { //所有比n大的都算一起，因为已经不可能是这个了，只计算有几篇
 8                 array[n] += 1
 9             }
10             else 
11             {
12                 array[citations[i]] += 1             //对应的次数+1
13             }
14         }
15         var t = 0
16         for i in (0...n).reversed() {
17             t = t + array[i] //累加次数
18             if(t >= i) {
19                 return i
20             }
21         }
22         return 0
23     }
24 }

40ms

 1 class Solution {
 2    func hIndex(_ citations: [Int]) -> Int {
 3         guard citations.count > 0 else {
 4             return 0
 5         }
 6         let res = citations.sorted()
 7         for (index,value) in res.enumerated() {
 8             if value >= (res.count - index){
 9                 return res.count - index
10             }
11         }
12     return 0
13     }
14 }

（编辑：李大同）

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