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?

和H-index类似，但是引用量已经排好序，需要在O(log n)的时间复杂度内解决问题。由于是有序数组，而且给定了O(log n)的时间复杂度，所以考虑采用二分法。对于任意一个下标i，我们可以确定它对应的引用量citation[i]，引用量大于等于citation[i]的论文数为n-i，那么我们利用二分法搜索下标i，使得它满足citation[i]>=(n-i)。

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