[LeetCode] 274. H-Index

[frdmu]

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, return compute the researcher's h-index.

According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.

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

Example 1:

Input: citations = [3,0,6,1,5]
Output: 3
Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 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, their h-index is 3.

Example 2:

Input: citations = [1,3,1]
Output: 1

计算H指数，维基百科中有解释，并给出了计算方法：

H指数的计算基于其研究者的论文数量及其论文被引用的次数。赫希认为：一个人在其所有学术文章中有N篇论文分别被引用了至少N次，他的H指数就是N。[1][2]如美国耶鲁大学免疫学家理查德·弗来沃发表的900篇文章中，有107篇被引用了107次以上，他的H指数是107。

可以按照如下方法确定某人的H指数：
     -  将其发表的所有SCI论文按被引次数从高到低排序；
     -  从前往后查找排序后的列表，只要当前的引用量大于当前的索引值，则H指数加1，最后得到的结果即为最终的H指数。

用图表示，如下所示：

代码如下：

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int h = 0;
        int N = citations.size();
        
        sort(citations.begin(), citations.end(), greater<int>());
        for (int i = 0; i < N; i++) {
            if (citations[i] > i)
                h++;
        }

        return h;
    }
};

有一说一，这题出的不咋地，好好的一个概念非得换成难以理解的说法，让人题都读的费劲。