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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 hindex=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nh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1n,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

posted on 2015-11-30 13:22  Coding菌  阅读(616)  评论(0编辑  收藏  举报