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.

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

 1 public class Solution {
 2     public int hIndex(int[] citations) {
 3         int[] res = new int[citations.length];
 4         mergesort(citations,res,0,citations.length-1);
 5         for(int i=0;i<citations.length;i++){
 6             if(i>=citations[i]){
 7                 return i;
 8             }
 9         }
10         return citations.length;
11     }
12     public void mergesort(int[] citations,int[] res,int start,int end){
13         if(start>=end) return;
14         int mid = start+(end-start)/2;
15         mergesort(citations,res,start,mid);
16         mergesort(citations,res,mid+1,end);
17         combine(citations,res,start,mid+1,end);
18     }
19     public void combine(int[] citations,int[] res,int left,int mid,int right){
20         int left_start  = left;
21         int left_end = mid-1;
22         int right_start = mid;
23         int right_end = right;
24         int k = left_start;
25         while(left_start<=left_end&&right_start<=right_end){
26             if(citations[left_start]>=citations[right_start]){
27                 res[k++] = citations[left_start++];
28             }else{
29                 res[k++] = citations[right_start++];
30             }
31         }
32         while(left_start<=left_end){
33             res[k++] = citations[left_start++];
34         }
35         while(right_start<=right_end){
36             res[k++] = citations[right_start++];
37         }
38         for(int i=left;i<=right;i++){
39             citations[i] = res[i];
40         }
41     }
42 }
43 //the run time complexity could be O(nlogn) ,the space complexity could be O(n);

 

posted @ 2017-05-25 08:54  CodesKiller  阅读(185)  评论(0)  编辑  收藏  举报