[LeetCode] H-Index

[LeetCode] 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 haveAt leastH citations each, and the other N −h papers haveNo moreH citations each ."

For example, given citations =[3, 0, 6, 1, 5], Which means the researcher has5Papers in total and each of them had received3, 0, 6, 1, 5Citations respectively. Since the researcher has3PapersAt least 3Citations each and the remaining twoNo more 3Citations each, his h-index is3.

Note: If there are several possible valuesh, The maximum one is taken as the h-index.

Solutions

Similar to counting sorting, slightly.

Implementation Code

C ++:

// Runtime: 4 msclass Solution {public:    int hIndex(vector
  
   & citations) {        int len = citations.size();        vector
   
     cnt(len + 1, 0);        for_each(citations.begin(), citations.end(),                 [len, &cnt](int n){ n >= len ? cnt[len]++ : cnt[n]++; });        for (int i = len; i >= 0; i--)        {            cnt[i] = i == len ? cnt[i] : cnt[i] + cnt[i + 1];            if (cnt[i] >= i)            {                return i;            }        }        return 0;    }};
   
  

Java:

// Runtime: 1 mspublic class Solution {    public int hIndex(int[] citations) {        int len = citations.length;        int cnt[] = new int[len + 1];        for (int c : citations) {            if (c >= len) {                cnt[len]++;            }            else {                cnt[c]++;            }        }        for (int i = len; i >= 0; i--) {            cnt[i] += i == len ? 0 : cnt[i + 1];            if (cnt[i] >= i) {                return i;            }        }        return 0;    }}