LeetCode 274 H-Index

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.

Hint:

1. An easy approach is to sort the array first.
2. What are the possible values of h-index?
3. A faster approach is to use extra space.

题意分析：

给出一个数组记录一个研究者各篇文章的引用数，写一个函数计算这个研究者的H指数。H指数是一个2005年由Jorge E. Hirsch提出的用于评估研究人员的学术产出数量与学术产出水平的指标。给定一个整数序列 citations=[3,0,6,1,5]，代表研究人员共有5篇论文，每个元素代表该论文的引用数量。从序列元素可以看出，例子  [3,0,6,1,5] ，很明显H的取值范围是0到5。该研究人员有至少3篇论文引用数量为>=3的，其余2篇论文引用数量不足3个引用，所以返回他的 h−index=3；
其实就是给你一个数组，找到一个最大的N,其中在数组中有N个元素大于等于N.

方法一：进行排序

	public int hIndex(int[] citations) {
		Arrays.sort(citations);
		int len = citations.length;
		for (int i = 0; i < len; i++) {
			if (citations[i] >= len - i)
				return len - i;
		}
		return 0;
	}

方法二：O（n），具体思路看代码。

	public int hIndex2(int[] citations) {
		int len = citations.length;
		int[] count = new int[len + 1];
		for (int citation : citations) {
			if (citation >= len) count[len]++;
			else count[citation]++;
		}
		int sum = 0;
		for (int i = len; i >= 0; i--) {
			sum += count[i];
			if (sum >= i) return i;
		}
		return 0;
	}