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.

count sort.

public class Solution {
    public int hIndex(int[] citations) {

        int[] cc = new int[citations.length+1];

        for(int i=0; i< citations.length; i++){
            if(citations[i]>= citations.length){
                cc[citations.length]++;
            }else{
                cc[citations[i]]++;
            }
        }

        int sum = 0;
        for(int i=cc.length-1; i>=0;i--){
            if(sum + cc[i] >= i){
                return i;
            }else
            {
                sum += cc[i];
            }
        }

        return 0;

    }
}

Sort the array in descending order, if i >= citations[i] then H-index is i;

public class Solution {
    public int hIndex(int[] citations) {

        Arrays.sort(citations);

        for(int i = citations.length - 1;  i>=0; i--){
            int h = citations.length - i-1;
            if(h >= citations[i]) return h;
        }

        return citations.length;

    }
}

sort the array, if citations[i], then there are citations.length - i of papers are > citations[i]; so the local H-index is min(citations[i], citations.length-i);

public class Solution {
    public int hIndex(int[] citations) {
        int h=0;
        Arrays.sort(citations);
        for(int i=0; i< citations.length; i++){
            h = Math.max(Math.min(citations[i], citations.length - i), h);
        }
        return h;
    }
}