問題描述：

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

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?

解題思路：

這道題目是h-index的一個follow up，是說給定數組已經排好序了怎麽做。

有要求時間復雜度在logn上。

顯然想到二分搜索。

先來找一下上界和下屆：從題目描述來看： 0 ≤ h ≤ N， 所以l = 0, r = n

然後來找移動的條件：

h-index的定義是：h篇論文的引用要至少為h，余下的(N-h)篇論文的引用都不能超過h

說明是引用數值和論文個數的關系。

求個數：n - i; n : citations.size() , i : 下標

求引用數：citations[i]

所以當n - i == citations[i]時： 代表：有citations篇論文的引用至少為citations[i], 余下的最多為citations[i],可以直接返回

若citations[i] < n-i 時：引用數小於論文篇數，所以我們應該增大引用數並減小論文數：left = i+1

若citations[i] > n-i 時：引用數大於論文篇數，所以我們應該增大論文數並減小引用數：right = i

最後n-left為h-index

代碼：

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

275. H-Index II