Given an array of citations sorted in ascending order (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."
Example:
Input:citations = [0,1,3,5,6]Output: 3 Explanation:[0,1,3,5,6]means the researcher has5papers in total and each of them had received 0, 1, 3, 5, 6citations respectively. Since the researcher has3papers with at least3citations each and the remaining two with no more than3citations each, her h-index is3.
Note:
If there are several possible values for h, the maximum one is taken as the h-index.
Follow up:
- This is a follow up problem to H-Index, where
citationsis now guaranteed to be sorted in ascending order. - Could you solve it in logarithmic time complexity?
H指数II。题意跟版本一差不多，本题已经帮你把input排好序了，并且题目要求时间复杂度需要在log级别，所以思路自然而然是二分法。
但是这个题如何做二分呢？首先因为input已经排好序，所以要比较的应该是citations[mid]和len - mid的关系。citations[mid]背后的含义是第mid篇文章被引用的次数；而len - mid的含义则要参考H指数的定义，
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.
所以还是参考这个图。如果citations[mid]正好等于len - mid则return mid；但是如果两者不等，则按照二分法正常的规则往中间逼近。注意最后返回的是len - left，代表h指数。
时间O(logn)
空间O(1)
Java实现
1 class Solution { 2 public int hIndex(int[] citations) { 3 int len = citations.length; 4 int left = 0; 5 int right = len - 1; 6 while (left <= right) { 7 int mid = left + (right - left) / 2; 8 if (citations[mid] == len - mid) { 9 return len - mid; 10 } else if (citations[mid] < len - mid) { 11 left = mid + 1; 12 } else { 13 right = mid - 1; 14 } 15 } 16 return len - left; 17 } 18 }
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