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 hcitations 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.

一刷
题解：首先创建一个数组，数组的index表示引用数目（0....>N), 数组内的元素表示文章数目。
然后从尾部开始，sum文章数目，直到sum>index则为我们所求的值。

public class Solution {
    public int hIndex(int[] citations) {
        int length = citations.length;
        if(length == 0) return 0;
        int[] array2 = new int[length+1];
        for(int i=0; i<length; i++){
            if(citations[i]>length)
                array2[length] += 1;
            else array2[citations[i]]+=1;
        }
        
        int t = 0, result = 0;
        for(int i=length; i>=0; i--){
            t += array2[i];//i indicate the citation, t indicate the paper number higher than the citation
            if(t>=i) return i;
        }
        return 0;
    }
}

二刷
题解：类似于统计词频， 用array[len+1]保存，array[len]表示引用>=len， array[i]表示引用次数为i的文章数目。
然后从尾部开始，sum， 如果sum>i, 则i为所求h

public class Solution {
    public int hIndex(int[] citations) {
        int length = citations.length;
        if (length == 0) {
            return 0;
        }
        
        int[] array2 = new int[length + 1];
        for (int i = 0; i < length; i++) {
            if (citations[i] > length) {
                array2[length] += 1;
            } else {
                array2[citations[i]] += 1;
            }
        }
        int t = 0;
        int result = 0;

        for (int i = length; i >= 0; i--) {
            t = t + array2[i];
            if (t >= i) {
                return i;
            }
        }
        return 0;
    }
}

三刷
同上

class Solution {
    public int hIndex(int[] citations) {
        int N = citations.length;
        int[] freq = new int[N + 1];
        for(int c : citations){
            if(c>=N) freq[N]++;
            else freq[c]++;
        }
        
        int sum = 0;
        for(int c=N; c>=0; c--){
            sum += freq[c];
            if(sum>=c) return c;
        }
        return 0;
    }
}