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.

h-index ，又称为h指数或h因子（h-factor），是一种评价学术成就的新方法。h代表“高引用次数”（high citations），一名科研人员的h指数是指他至多有h篇论文分别被引用了至少h次, 剩余的论文引用的次数不多于h次。题目中给定了一个数组，数组的长度就代表了论文的个数，每个元素的值代表了引用的次数，让我们从中找到H-index.

首先我们设定h的初始值为1。然后我们将数组排序，从最后一个元素开始与h比较，1，如果比h大，就让h加1，继续往前比较，直到遇到等于或者小于h的情况；2， 如果等于h, 就返回h; 3，如果小于h, 就返回h - 1。在第1步中，如果一直都大于h, 我们就返回数组的长度。我们用到了排序，这样时间复杂度为O(nlogn)。代码如下：

public class Solution {
    public int hIndex(int[] citations) {
        if(citations == null || citations.length == 0) return 0;
        Arrays.sort(citations);
        int h = 1;
        for(int i = citations.length - 1; i >= 0; i--) {
            if(citations[i] > h) {
                h ++;
            } else if(citations[i] == h) {
                return h;
            } else {
                return h - 1;
            }
        }
        return citations.length;
    }
}

另外一种方法只用O(n)的时间复杂度就可以解决，但是我们要用到O(n)的空间。类似于计数排序的方法，创建一个数组count来记录文章引用次数的情况，然后从count的最后一个元素开始累加，当累加的数字大于或等于当前下标的时候就返回。代码如下：

public class Solution {
    public int hIndex(int[] citations) {
        if(citations == null || citations.length == 0) return 0;
        int len = citations.length;
        int[] count = new int[len + 1];
        for(int i = 0; i < len; i++) {
            if(citations[i] >= len) 
                count[len] ++;
            else
                count[citations[i]] ++;
        }
        for(int i = count.length - 1; i > 0; i--) {
            if(count[i] >= i) return i;
            count[i - 1] += count[i];
        }
        return 0;
    }
}