题目

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.

分析

题目描述的要求如下：

给定一个整数序列 citations=[3,0,6,1,5]，代表研究人员共有5篇论文，每个元素代表该论文的引用数量。从序列元素可以看出，该研究人员有至少3篇论文引用数量为>=3的，其余2篇论文引用数量不足3个引用，所以返回他的 h−index=3；

也就是说，我们找返回一个整数h，使得数组中至少h个元素值大小>=h，其n−h个元素值<h。

解决方法：

首先对序列排序，然后从大到小遍历数组，h值为从1到n，若元素满足num[i]>h，继续遍历，否则跳出循环，返回h即可。

AC代码

class Solution {

public:

    int hIndex(vector<int>& citations) {

        if (citations.empty())

            return 0;

        //对所给序列排序

        sort(citations.begin(), citations.end());

        int len = citations.size(),maxH = 0;

        for (int i = len - 1; i >= 0; --i)

        {

            int h = len - i;

            if (citations[i] >= h && h > maxH)

            {

                maxH = h;

            }

            else{

                break;

            }

        }//for

        return maxH;

    }

};

GitHub测试程序源码