LeetCode #274 - H-Index-优快云博客

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/81193709

本文介绍了一种计算研究人员h指数的方法。h指数是指一个科学家有h篇论文至少被引用了h次，其余的论文每篇的引用次数不超过h次。通过统计不同引用次数的论文数量并遍历这些数量，我们可以找出最大的h值。

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题目描述：

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."

Example:

Input: citations = [3,0,6,1,5]Output: 3 
Explanation: [3,0,6,1,5] 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, her h-index is 3.

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

给定一个数组，表示多篇论文的引用数量，那么h系数的定义为最少有h篇论文的引用数量不少于h次，求这个h系数。由于引用次数可能相同，所以需要先统计每个引用量对应的文章数，当引用量大于n，就记为n次。那么我们从n个引用量开始遍历，可以确定一共有多少篇论文的引用量大于等于n，然后一直向前遍历直到满足论文数大于等于引用量。

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n=citations.size();
        vector<int> v(n+1,0);
        for(int i=0;i<n;i++)
        {
            if(citations[i]<=n) v[citations[i]]++;
            else v[n]++;
        }
        int count=0;
        int i=n;
        while(i>=0)
        {
            count+=v[i];
            if(count<i) i--;
            else if(count>=i) return i;
        }
    }
};