LeetCode #275 H-Index II

Question

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.

二分查找O(logn)

这题表面上是承接了H-index的一道题，实际上是一道典型的Binary Search题。

当citations排好序之后，直接找出citations[i] < i的位置即可。由于这里上升序排序，所以实际写的时候和降序稍有不同。

但此题的关键不是想到用Binary Search，而是处理Binary Search那些复杂的细节。这里介绍C++标准库<algorithm>里的超简洁、bug free的通用写法：

def lower_bound(array, first, last, value): # 返回[first, last)内第一个不小于value的值的位置
    while first < last: # 搜索区间[first, last)不为空
        mid = first + (last - first) // 2  # 防溢出 
        if array[mid] < value: first = mid + 1 
        else: last = mid 
　　 return first  # last也行，因为[first, last)为空的时候它们重合

这样写的好处很多：

1. 简洁好记，只有first = mid+1 处用到了位置调整

2. 即使区间为空、答案不存在、有重复元素、搜索开/闭的上/下界也同样适用，鲁棒性高

要点：

1. 搜索区间是左闭右开，[first, last)

2. 只有first = mid + 1，而last = mid （这能避免死循环，详见参考资料）

3.  mid = first + (last - first) // 2 （防溢出，适用于指针、迭代器，当然python大数可以不用考虑溢出）

最后将这个方法应用在本题中：

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        length = len(citations)
        left = 0
        right = length
        while left < right:
            mid = left + (right - left) // 2
            if citations[mid] < length - mid: left = mid + 1
            else: right = mid
        return length - left

非常简洁

python中其实也有二分查找的实现 bisect，见参考资料

参考：

https://www.zhihu.com/question/36132386/answer/530313852

https://docs.python.org/2/library/bisect.html