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# H-Index II(二分）

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](https://en.wikipedia.org/wiki/H-index): "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](https://leetcode.com/problems/h-index/description/)

  , where

  `citations`

  is now guaranteed to be sorted in ascending order.
* Could you solve it in logarithmic time complexity?

分析

用标准二分模板做。记得条件是num\[m]>=n-m（长度）

H-index I完全一样，就是需要排序而已

```
class Solution:
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """

        n = len(citations)
        if n == 0:
            return 0
        s, e = 0, n - 1
        while s + 1 < e:
            m = s + (e - s) // 2

            if n-m <= citations[m]:
                e = m
            else:
                s = m
        if n-s <= citations[s]:
            return n-s
        if n-e <= citations[e]:
            return n-e
        return 0
```