LeetCode 275. H-Index II – 二分查找（Binary Search）系列题16

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in an ascending order, return compute the researcher's h-index.

According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.

If there are several possible values for h, the maximum one is taken as the h-index.

You must write an algorithm that runs in logarithmic time.

Example 1:

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, their h-index is 3.

Example 2:

Input: citations = [1,2,100]
Output: 2

Constraints:

• n == citations.length
• 1 <= n <= 105
• 0 <= citations[i] <= 1000
• citations is sorted in ascending order.

题目里又是有序数组又是O(logn)时间复杂度的，相当于直接告诉我们要用二分查找法。仔细分析题目可以把题目简化为：给定一个长度为n的有序数组citations，从[1,n]范围内找一个最大数h，使得数组里至少有h个数的数值大于等于h。这样用二分查找法已经很明显且很容易了。

解题思路：把长度为n的数组citations从中点mid = (l + r) // 2 分成左右两个区间，中点数的值是citations[mid]，由于数组是递增的，那么从mid开始到结尾都是大于等于中点数值citations[mid]的，个数为n - mid, 如果citations[mid]大于等于n-mid，那么说明存在一个h = n - mid使得数组里至少存在h个数的数值大于等于h，因此可以试图查找更长的h, 也就是继续在左半区间里查找。如果citations[mid]小于n-mid，左半区间的数值肯定都小于h=n-mid，则大于等于h = n - mid的个数肯定少于h个，因此需要在右半区间查找更短一点的h。

基于以上的思路就很容易实现二分查找的算法了。

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