[Swift]LeetCode275. H指数 II | H-Index II

Given an array of citations sorted in ascending order (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 0 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 = [0,1,3,5,6][0,6]5,63333

Note:

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

Follow up:

• This is a follow up problem to H-Index,where citations is Now guaranteed to be sorted in ascending order.
• Could you solve it in logarithmic time complexity?

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），数组已经按照升序排列。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 0 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 = [0,63333

说明:

如果 h 有多有种可能的值 ，h 指数是其中最大的那个。

进阶：

• 这是 H指数 的延伸题目，本题中的 citations 数组是保证有序的。
• 你可以优化你的算法到对数时间复杂度吗？

208ms

 1 class Solution {
 2     func hIndex(_ citations: [Int]) -> Int {
 3         let count = citations.count
 4         var left = 0,right = count - 1
 5         while left <= right {
 6             let mid = (left + right) / 2
 7             if citations[mid] == count - mid {
 8                 return count - mid
 9             }else if citations[mid] > count - mid {
10                 right = mid - 1
11             }else {
12                 left = mid + 1
13             }
14         }
15         
16         return count - left
17     }
18 }

232ms

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