[Swift]LeetCode275. H指数 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: citations = [0,1,3,5,6]
Output: 3 
Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had 
             received 0, 1, 3, 5, 6 citations respectively. 
             Since the researcher has 3 papers with at least 3 citations each and the remaining 
             two with no more than 3 citations each, her h-index is 3.

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: citations = [0,1,3,5,6]
输出: 3 
解释: 给定数组表示研究者总共有 5 篇论文，每篇论文相应的被引用了 0, 1, 3, 5, 6 次。
     由于研究者有 3 篇论文每篇至少被引用了 3 次，其余两篇论文每篇被引用不多于 3 次，所以她的 h 指数是 3。

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

class Solution {
    func hIndex(_ citations: [Int]) -> Int {
        let count = citations.count
        var left = 0, right = count - 1
        while left <= right {
            let mid = (left + right) / 2
            if citations[mid] == count - mid {
                return count - mid
            }else if citations[mid] > count - mid {
                right = mid - 1
            }else {
                left = mid + 1
            }
        }
        
        return count - left
    }
}

---

232ms

class Solution {
    func hIndex(_ citations: [Int]) -> Int {
        guard citations.count > 0 else {
            return 0
        }
        for (index,value) in citations.enumerated() {
            if value >= (citations.count - index){
                return citations.count - index
            }
        }
        return 0
    }
}