Leetcode H-index

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.

思路：  先将citations排序，假设case1:  [0, 1, 3, 4, 5, 5, 5]，这时正好有4篇引用数至少为4的文章，所以H-index就是4。这一题实际上是在search第一个不满足条件（ (n - i) > citat[i] ）

如果case2: [0, 1, 3, 5, 5, 5, 5]，那么第一个不满足条件的位置是 citat[3] = 5，因为引用数大于等于5的文章只有n-3 = 4篇。这时的H-index就是n-3 = 4。

再例如case3: [100, 100]，第一个不满足条件的位置是0，因为显然没有100篇文章被至少引用了100次。所以H-index = n - 0 = 2

注意L21，因为如果发生了case2或者case3，最后的 right 会由于left = right时不满足条件而运行L21，导致right跑到left左边去。所以最后停在第一个不满足条件的index上的是left。

 class Solution(object):
     def hIndex(self, citations):
         """
         :type citations: List[int]
         :rtype: int
         """
         if not citations:
             return 0

         n = len(citations)
         left = 0
         right = n - 1

         while left <= right:
             i = left + (right - left)/2
             if citations[i] == n-i:
                 return n - i
             elif citations[i] < n - i:
                 left = i + 1
             else:
                 right = i - 1

         return n - left