[LeetCode]H-Index

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本文介绍了一种计算研究人员h指数的方法。h指数是指在N篇论文中，有h篇被引用次数不低于h次，其余的N-h篇论文的引用次数不超过h次。文章通过排序和遍历数组的方式实现了这一计算过程。

Given an array of citations (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."

For example, given citations = [3, 0, 6, 1, 5], which means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 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, his h-index is 3.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

题目意思比较绕，意思是：

一个人有N篇文章，每篇文章有若干引用，index 是这样一个数字，他所有文章中有h篇（0<＝h<＝N)，的引用数不小于h，其他N-h不大于h。

先对数组排序，然后对h遍历，那么citations[i]表示当前引用数，那么大于citations[i]的就有 size()-i 篇，小于等于其的就有 i篇。当citations[i]>=size()-i时表示，当时的i时可用用做h的。

class Solution {
    public:
    int hIndex(vector<int>& citations) {
    
        sort(citations.begin(),citations.end());
    
        int size = citations.size();
        if(size == 0){
            return 0; 
        }
    
        int h_index = 0;
        for(int i = 0; i<size;i++){
            if(citations[i] >= (size-i) ){
                h_index = size-i;
                return h_index;
            }
        }
        return h_index;
    
    }
};