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

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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][0,1,3,5,6]5, 1, 3, 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?

---

給定一位研究者論文被引用次數的數組（被引用次數是非負整數），數組已經按照升序排列。編寫一個方法，計算出研究者的 h 指數。

h 指數的定義: 「h 表明「高引用次數」（high citations），一名科研人員的 h 指數是指他（她）的 （N 篇論文中）至多有 h 篇論文分別被引用了至少 h 次。（其他的 N - h 篇論文每篇被引用次數很少於 h 次。）"

示例:

輸入: 
輸出: 3 
解釋: 給定數組表示研究者總共有  篇論文，每篇論文相應的被引用了 0 次。
     因爲研究者有 篇論文每篇至少被引用了  次，其他兩篇論文每篇被引用很少於  次，因此她的 h 指數是 。citations = [0,1,3,5,6]5, 1, 3, 5, 63333

說明:

若是 h 有多有種可能的值 ，h 指數是其中最大的那個。

進階：

• 這是 H指數 的延伸題目，本題中的 citations 數組是保證有序的。
• 你能夠優化你的算法到對數時間複雜度嗎？

---

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 }