[LeetCode#274]H-Index

Problem:

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.

Analysis:

This problem is interesting!!! It tests your coding skill and logic ability.

Since h-index defines that at least h papers should exceed(include) h citation, you may wrongly think this is a simple count problem, why not use HashMap<citation, count>. 
However, even paper that have citation' larger than citation should be counted! What a pity, Right?
Apparently the HashMap should not be used!!!

Since there is good prperty : all papers have citation exceed certain index, could be counted for that index. Why not use sort??? Then, sorting the citation array in descending order, (i+1) is the total citations exceed citation[i]. 
Then I have following implementations:

Wrong solution 1:

public class Solution {
    public int hIndex(int[] citations) {
        if (citations == null)
            throw new IllegalArgumentException("The citaions' reference is null!");
        Arrays.sort(citations, Collections.reverseOrder());
        for (int i = 0; i < citations.length; i++) {
            if (i+1 >= citations[i])
                return citations[i];
        }
        return citations.length;
    }
}
Error:
Line 5: error: no suitable method found for sort(int[],Comparator<Object>)
Mistake 1:
Arrays.sort(citations, Collections.reverseOrder()); 
Not work for primitive type, it only works for Integer, Double ....

Wrong solution 2:

Since we should give up the way of sorting citations in descending order, we should just use ascending order. 
For the citation in ascending order, citation[i]'s useful count is the number of papers after it (inclusive).
for (int i = 0; i < citations.length; i++) {
    if (citations.length - i >= citations[i])
        ...
}


public class Solution {
    public int hIndex(int[] citations) {
        if (citations == null)
            throw new IllegalArgumentException("The citaions' reference is null!");
        Arrays.sort(citations);
        int max = -1;
        for (int i = 0; i < citations.length; i++) {
            if (citations.length - i >= citations[i])
                max = Math.max(max, citations[i]);
        }
        return (max == -1 ? citations.length : max);
    }
}

Errors:
Input:
[4,4,0,0]
Output:
0
Expected:
2

However, the above solution only consider the situation of "citations.length - i >= citations[i]" and no citation[i] is valid case (at the end).
Even we may not be able to find citations[i] meet:
if (citations.length - i >= citations[i])
    max = Math.max(max, citations[i]);

We still should have a valid h-index!
Suppose we have no "citations.length - i >= citations[i]" case, it means
citations.length - i < citations[i]

since "citations[i]"" < "citations[citaions.length - i]"(thus all citations[i] account into citations[citaions.length - i]).
Thus we must have (possible)hindex = citations.length - i.

Solution:

public class Solution {
    public int hIndex(int[] citations) {
        if (citations == null)
            throw new IllegalArgumentException("The citaions' reference is null!");
        Arrays.sort(citations);
        int max = 0;
        for (int i = 0; i < citations.length; i++) {
            if (citations.length - i >= citations[i])
                max = Math.max(max, citations[i]);
            else 
                max = Math.max(max, citations.length - i);
        }
        return max;
    }
}