Leetcode notes: H-index

I. Title Description

Given an array of citations (each citation was a non-negative integer) of a researcher, write a function to compute the RES Earcher ' s h-index.

According to the definition of H-index on Wikipedia: ' A scientist has index h if h of his/her N papers having at least h CIT Ations, and the other N? H Papers has no more than H citations each. "

For example, given citations = [3, 0, 6, 1, 5] , which means the researcher have 5 papers in total and each of the them had received 3, 0, 6, 1, 5 citations R Espectively. Since the researcher have 3 papers with at least 3 citations each and the remaining both with no more than 3 citations each, His h-index is 3.

Note:if there was several possible values for H, and the maximum one is taken as the h-index.

Two. Topic analysis

First you need to understand the main idea of the topic:

Given an array, the number of citations of a researcher is documented (the number of citations for each article is a nonnegative integer), and the writing function calculates the H index of the researcher.

According to Wikipedia's definition of the H index: "The H index of a scientist means that in his published N paper, a paper is h quoted at least h once, and N-h the number of citations for the remainder of the article is no more than h ".

For example, given an array citations = [3, 0, 6, 1, 5] , this means that the researcher has a total of 5 papers, each of which gets a 3, 0, 6, 1, 5 secondary reference. Since the researchers had 3 at least one citation, the 3 other two citations did not exceed the number of 3 times, so the H index was 3 .

Note: If there are more than one possible h value, take the maximum value as an h exponent.

Through, you can more intuitively understand the h definition of the value, corresponding to the graph, that is, the lower left corner of the ball square the maximum value:

The following explanation assumes that the size of the given array is N , that is, a total N article.

There are two general practices, also referred to in the topic tips, the first thought is to sort the array, and then traverse from the back to find the H value, the complexity of the method is: O(n*logn) .

In the interview, if you allow the use of secondary memory, you can use the second method, that is, to open up a new array record , to record the 0~N number of times cited several articles (the number of references greater than N the N second calculation) to iterate over the array, after the statistics, to iterate over the statistical array record , hthe maximum value of the value can be calculated. The complexity of time is O(n) .

Three. Sample code

//sort + traverse  class  Solution {public : int  hindex (vector  <int  >  & citations) {sort (Citations.begin (), Citations.end (), [] (const  int  &a, const  int  &b) {return  a > B;}); int  i = 0 ; for  (; i < citations.size (); ++i) if  (Citations[i] <= i) break ; return  i; }};

//The second methodclassSolution { Public:intHindex ( vector<int>& citations) {intCitationsize = Citations.size ();if(Citationsize <1)return 0; vector<int>Record (Citationsize +1,0); for(inti =0; i < citationsize; ++i) {if(Citations[i] <= citationsize) ++record[citations[i];Else++record[citationsize]; } for(intj = citationsize, Papernum =0; J >=0; --J) {papernum + = Record[j];if(Papernum >= J)returnJ }return 0; }};

Four. Summary

Which method to use depends on the actual conditions.

Leetcode notes: H-index