H-Index

描述

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.

分析

H-Index的含义是,如果一个人发表的所有论文中,有h篇论文分别被引用了至少h次,那么他的H-Index就是h

思路一:先从大到小排序,然后从前往后扫描,如果当前文章数(即当前下标+1)等于值本身,则返回当前文章数作为 h-index;如果当前文章数大于值本身,则返回当前文章数-1作为H-Index, 因为当前文章的引用数小于当前文章数,不能算在内。时间复杂度O(nlognO(n\log{}n,空间复杂度O(1)。

思路二:跟思路一类似,不过排序算法换成了计数排序。有一个小技巧,因为H-Index最大不可能超过论文综述,所以我们只需要开一个长度为n+1的数组,如果某篇论文的引用数超过了n,就将其当做n

代码1 全排序

  1. // H-Index
  2. // Time complexity: O(nlogn), Space complexity: O(1)
  3. public class Solution {
  4. public int hIndex(int[] citations) {
  5. Arrays.sort(citations);
  6. reverse(citations);
  7. for (int i = 0; i < citations.length; ++i) {
  8. if (i + 1 == citations[i]) return i+1;
  9. if (i + 1 > citations[i]) return i;
  10. }
  11. return citations.length;
  12. }
  13. private static void reverse(int[] nums) {
  14. int left = 0;
  15. int right = nums.length - 1;
  16. while (left < right) {
  17. final int tmp = nums[left];
  18. nums[left] = nums[right];
  19. nums[right] = tmp;
  20. ++left;
  21. --right;
  22. }
  23. }
  24. }

代码2 计数排序

  1. // H-Index
  2. // Time complexity: O(n), Space complexity: O(n)
  3. public class Solution {
  4. public int hIndex(int[] citations) {
  5. final int n = citations.length + 1;
  6. final int[] histogram = new int[n+1];
  7. for (int x : citations) {
  8. ++histogram[x > n ? n : x];
  9. }
  10. int sum = 0; // current number of papers
  11. for (int i = n; i > 0; --i) {
  12. sum += histogram[i];
  13. if (sum >= i) {
  14. return i;
  15. }
  16. }
  17. return 0;
  18. }
  19. }

相关题目

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/sorting/counting-sort/h-index.html