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

概述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-

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 - 111             }else {12                 left = mID + 113             }14         }15         16         return count - left17     }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 012     }13 }