<LeetCode OJ> 274 / 275 H-Index(I / II)

最新推荐文章于 2025-08-09 17:01:50 发布

EbowTang

最新推荐文章于 2025-08-09 17:01:50 发布

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分类专栏： LeetCode解题报告文章标签： leetcode 面试 C++

本文链接：https://blog.youkuaiyun.com/EbowTang/article/details/50849690

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本文介绍了一种计算研究人员h指数的方法，包括未排序和已排序两种情况下的算法实现。通过排序和二分查找的方式，有效地找到了满足条件的h指数。

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Total Accepted: 37719 Total Submissions: 125936 Difficulty: Medium

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：

本题比较简单，不多说！

class Solution {
public:
    int hIndex(vector<int>& citations) {
        sort(citations.begin(),citations.end());//排序
        for(int i=0;i<citations.size();i++)
            if(citations[i] >= (citations.size()-i))//如果当前引用数大于它的位置（倒着数），那么就是他了
                return (citations.size()-i);
        return 0;        
    }
};

Total Accepted: 25892 Total Submissions: 78747 Difficulty: Medium

Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize your algorithm?

Hint:

Expected runtime complexity is in O(log n) and the input is sorted.

分析：

class Solution {
public:
    int hIndex(vector<int>& citations) {
        if (citations.empty()) 
            return 0;
        int start = 0, len = citations.size(), end = len - 1;
        while (start <= end) {
            int mid = (start + end)/2;
            if (citations[mid] < len - mid)
                start = mid + 1;
            else if (citations[mid] > len - mid)
                end = mid - 1;
            else 
                return len - mid;
        }
        return len - start;
    }
};

另外的写作形式：

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int len = citations.size(), low = 0, height = len-1, mid;  
        while (low <= height) {  
            mid = (low + height) / 2;  
            if (citations[mid] >= len - mid)  
                height = mid-1;  
            else 
                low = mid + 1;  
        }  
        return len - low; 
    }
};

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.youkuaiyun.com/ebowtang/article/details/50849690

原作者博客：http://blog.youkuaiyun.com/ebowtang

本博客LeetCode题解索引：http://blog.youkuaiyun.com/ebowtang/article/details/50668895