275. H-Index II

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?

  public class Solution {
      public int hIndex(int[] citations) {
          final int n = citations.length;
          int begin = 0;
          int end = citations.length;
  
          while (begin < end) {
              final int mid = begin + (end - begin) / 2;
              if (citations[mid] < n - mid) {
                  begin = mid + 1;
              } else {
                  end = mid;
              }
          }
          return n - begin;
      }
  }

  设数组长度为n，那么n-i就是引用次数大于等于nums[i]的文章数。如果nums[i]<n-i，说明i是有效的H-Index, 如果一个数是H-Index，那么最大的H-Index一定在它后面（因为是升序的），根据这点就可以进行二分搜索了。