Leetcode - 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.

[分析]
扫描一遍citations，使用一个额外数组count[]计数每个引用数的文章数量，引用数大于N（文章总数）的视作N。然后从后往前扫描count[]，同时累加已扫描部分记为sum，sum表示引用数>=i的文章总数，因为count[]是从后往前扫描，发现的第一个i即为所求的h-index。

public class Solution {
    public int hIndex(int[] citations) {
        if (citations == null || citations.length == 0)
            return 0;
        int N = citations.length;
        int[] count = new int[N + 1]; // count[i] means number of articles which citations >= i
        for (int i = 0; i < N; i++) {
            if (citations[i] > N)
                count[N] += 1;
            else
                count[citations[i]] += 1;
        }
        int sum = 0;
        for (int i = N; i >= 0; i--) {
            sum += count[i];
            if (sum >= i)
                return i;
        }
        return 0;
    }
}