LeetCode:H-Index

H-Index

Total Accepted: 34313 Total Submissions: 115907 Difficulty: Medium

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.

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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 (M) H-Index II

思路：

首先阐述一个事实：h-index不超过L（即h-index <= L，L是citations数组长度）；当h-index == L时，citations所有值都不小于L；

创建一个counts数组，存储citation == index（index<L时），当index>=L时，存入counts[L]；

如有数组citations = [3,0,7,1,5]，L = 5；

创建counts[L+1]数组，存储方式为：

if(citation == index) counts[index]++;

if(citation >= L) counts[L]++;

最后，从右向左遍历counts，并用ans变量记录个数；当ans>=i时，即得解。

java code:

public class Solution {    public int hIndex(int[] citations) {                int L = citations.length;                int[] counts = new int[L+1];                for(int i:citations) {            if(i>=L) counts[L]++;            else counts[i]++;        }                int ans = 0;        for(int i=L;i>=0;i--) {            ans += counts[i];            if(ans>=i) return i;        }        return 0;    }}