[LeetCode]274. H-Index（自学留存）-优快云博客

本文链接：https://blog.youkuaiyun.com/Ada_who/article/details/79362724

本文介绍了一种高效计算研究人员h指数的方法。h指数是一种衡量科学家影响力的标准，它基于作者出版物的引用次数来评估其学术成就。文章提出了一种改进算法，通过构建一个特殊数组来追踪不同引用数量的文章数目，从而避免了直接排序导致的时间复杂度过高问题。

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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."

For example, given citations = [3, 0, 6, 1, 5], which 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, his h-index is 3.

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

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Personal tips: 一开始使用排序后操作，但是时间复杂度太差。可以新建一个数组a[被引用次数]=该被引用的篇数。其中被引用次数大于等于总篇数均记录在总篇数s里，即a[s]=大于等于s的篇数。代码如下：

class Solution {
public:
	int hIndex(vector<int>& citations) {
		if (citations.empty()) return 0;
		int s = citations.size();
		vector<int> times(s+1, 0);
		for (auto c : citations) ++times[min(c,s)];
		int index = 0;
		for (int i = s;i>=0;i--)
		{
			index += times[i];
			if (index >= i) return i;
		}
		return 0;
	}
};