leetcode 274. H-Index 高引用次数计算1

柔光的暖阳◎ 2022-06-08 13:25 155阅读 0赞

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.

这个直接根据H-Index的百科即可。

代码如下：

import java.util.Arrays;
    
    /*
     * https://baike.baidu.com/item/h-index/3991452?fr=aladdin
     * 参见H-Index的百科即可
     * */
    class Solution
    {
        
        public int hIndex(int[] citations)
        {
            if(citations==null || citations.length<=0)
                return 0;
    
            Arrays.sort(citations);
            int maxH=0;
            for(int i=citations.length-1;i>=0;i--)
            {
                int count=citations.length-i;
                if(citations[i]>=count && count>maxH)
                    maxH=count;
                else
                    break;
            }
            return maxH;
        }
    }

下面是C++的做法，按照H-Index的含义去做即可，代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <algorithm>
    
    using namespace std;
    
    class Solution 
    {
    public:
        int hIndex(vector<int>& c) 
        {
            sort(c.begin(),c.end());
            int maxH = 0;
            for (int i = c.size() - 1; i >= 0; i--)
            {
                int count = c.size() - i;
                if (c[i] >= count && count > maxH)
                    maxH = count;
    
            }
            return maxH;
        }
    };