[Leetcode]H-Index

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篇论文至少被引用了h次,而其余的N-h篇论文最多被引用了h次。文章提供了两种算法实现:一种是时间复杂度为O(nlogn)的排序方法;另一种是时间复杂度为O(n)的计数排序方法。

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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.

Hint:

  1. An easy approach is to sort the array first.
  2. What are the possible values of h-index?
  3. A faster approach is to use extra space.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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class Solution {
public:
    /*algorithm:
        time O(nlogn) space O(1)
    */
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        sort(citations.begin(),citations.end(),greater<int>());
        for(int i = 0;i < n;i++){
            if(i+1 > citations[i])return i;
        }
        return n;
    }
};

class Solution {
public:
    /*algorithm: count sort
        time O(n) space O(n)
        i+1 > citations
    */
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        vector<int>cnt(n+1,0);
        for(int i = 0;i < n;i++){
            cnt[min(n,citations[i])]++;
        }
   
        int sum = 0;
        for(int i = n;i >= 0;i--){
            sum += cnt[i];
            if(sum >= i)return i;
        }
    }
};


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