LeetCode Q274: 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 thanh 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.

Hint:

1. An easy approach is to sort the array first.
2. What are the possible values of h-index?
3. A faster approach is to use extra space.

Solution 1: Sort

	class Solution {
	public:
	int hIndex(vector<int>& citations) {
	if(citations.size()==0)
	return 0;
	sort(citations.begin(), citations.end());
	reverse(citations.begin(), citations.end());
	int h=0;
	for(int i=0; i<citations.size(); i++){
	int tmph;
	tmph = i<citations[i]? i+1:citations[i];
	h=max(tmph, h);
	}
	return h;
	}
	};

view raw Q274-1.cpp hosted with ❤ by GitHub

Solution 2: Hash table
The idea is lay in the fact that the h-index can not be larger than the length of the array. So, we can use a array with length n to store the number of papers with citation equal to the index of the array.

	int hIndex(vector<int>& citations) {
	if(citations.empty())
	return 0;
	int n = citations.size();
	vector<int> hash(n + 1, 0);
	for(int i = 0; i < n; ++i){
	if(citations[i] >= n)
	hash[n]++;
	else
	hash[citations[i]]++;
	}
	int paper = 0;
	for(int i = n; i >= 0; --i){
	paper += hash[i];
	if(paper >= i)
	return i;
	}

view raw Q274-2.cpp hosted with ❤ by GitHub