LeetCode: 275. H-Index II

题目描述

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?

解题思路 —— 二分查找

• citations[mid] >= citations.size()-mid: 尝试在 ‘[left, mid)’ 寻找使得 h 更大的 mid
• 否则, 尝试在 ‘[mid, right)’ 寻找符合题意的 mid

AC 代码

class Solution 
{
public:
    int hIndex(vector<int>& citations) 
    {
        int left = 0, right = citations.size();
        while(left < right)
        {
            int mid = (left+right)/2;
            if(citations[mid] >= citations.size()-mid)
            {
                // [left, mid)
                right = mid;
            }
            else
            {
                // [mid+1, right)
                left = mid + 1;
            }
        }
        
        return citations.size()-left;
    }
};