【刷题-LeetCode】275. H-Index II

1. H-Index II

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 h-index：和爱丁顿数一个定义。对于一个从小到大排列好的数组，h-index为：

\[h = \arg \max \{nums[n-i] >= i\}
\]

最简单是线性搜索，逐一判断即可。在有序数组中，可以使用二分查找

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