LeetCode 275. H-Index II

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 has 5 papers in total and each of them had 
             received 0, 1, 3, 5, 6 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, her h-index is 3.

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.