原题链接在这里:https://leetcode.com/problems/h-index-ii/
题目:
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 has5
papers in total and each of them had
received 0, 1, 3, 5, 6
citations respectively.
Since the researcher has3
papers with at least3
citations each and the remaining
two with no more than3
citations 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
citations
is now guaranteed to be sorted in ascending order. - Could you solve it in logarithmic time complexity?
题解:
Have l = 0, r = n - 1, continue binary search with l <= r, if (nums[mid] == n - mid), we find the target.
Otherwise, when nums[mid] < n - mid, that means we guess smaller, l = mid+ 1. Else, r = mid - 1.
After gettting out of binary search and there is no result, return n - l.
Note: here it is n - l, not n - 1.
Time Complexity: O(logn).
Space: O(1).
AC Java:
public class Solution {
public int hIndex(int[] citations) {
if(citations == null || citations.length == 0){
return 0;
}
int len = citations.length;
int r = len-1;
int l = 0;
while(l<=r){
int mid = l+(r-l)/2;
if(citations[mid] == len-mid){
return len-mid;
}else if(citations[mid] < len-mid){
l = mid+1;
}else{
r = mid-1;
}
}
return len-l;
}
}
类似H-Index.