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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.
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();
}
};
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:
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:
类似题目:
H-Index II
参考资料:
https://leetcode.com/problems/h-index/
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