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.”
给一个升序数组，代表一个研究人员每篇文章的引用次数，求这个研究人员的H指数。H指数代表这个研究人员至少有H篇文章被至少引用了H次，返回所有可能H值中最大的一个。

样例

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.

算法1

题解：首先这个人的H指数肯定是在$[0,nums.size()]$之间的，这个题目可以看作是求所有H值中最大的那个，但是我们可以反向思考一下。如下:

index    = [0,1,2,3,4]//数组下标
citation = [0,1,3,5,6]//citation[i]代表第i篇论文的引用次数
count    = [5,4,3,2,1]//count[i]代表至少有多少篇文章引用了引用了citation[i]次
其他例子
// [0,0,4,4]
// [4,3,2,1]

// [0,0,0,0]
// [4,3,2,1]

// [5,6,7,8]
// [4,3,2,1]

那么问题其实可以转化成求所有$citation[i] >= count[i]$中最小的那个$i$对应的$count[i]$，这样问题就转换成二分查找模板二 的问题了，求使条件成立的最小的$i$。如果没有任何一个$i$满足，则说明这个人的H值应该为0。这里count[i] = n - i。

数组是升序的可以直接判断最后一个元素如果是0，那么返回0，否则至少有一篇文章引用了被引用过，$H \geq 1$

C++ 代码

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