1352. Product of the Last K Numbers

 

Given an array of citations (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 = [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, her h-index is 3.

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

 

这道题让我们求H指数，这个质数是用来衡量研究人员的学术水平的质数，定义为一个人的学术文章有n篇分别被引用了n次，那么H指数就是n。而且wiki上直接给出了算法，可以按照如下方法确定某人的H指数：1、将其发表的所有SCI论文按被引次数从高到低排序；2、从前往后查找排序后的列表，直到某篇论文的序号大于该论文被引次数。所得序号减一即为H指数。我也就没多想，直接按照上面的方法写出了代码：

 

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

 

类似题目：

H-Index II

 

参考资料：

https://leetcode.com/problems/h-index/

 

LeetCode All in One 题目讲解汇总(持续更新中…)

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