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LeetCode 274. H指数（排序，哈希）

先排序，然后找到最后一个可形成正方形的点 class Solution { public: int hIndex(vector& citations) { sort(citations.begin...(), citations.end(), [](int &a, int &b){return a > b;}); int i = 0; while(i citations.size...二分法 class Solution { public: int hIndex(vector& citations) { sort(citations.begin(),...citations.end(), [](int &a, int &b){return a > b;}); //找最后一个citations[i] >= i+1 int...if(citations[mid] < mid+1) r = mid-1; else { if(mid == citations.size

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Baozi Training Leetcode solution 274: H-Index

1 public int hIndex(int[] citations) { 2 if (citations == null || citations.length == 0) { 3...] 7 int[] counts = new int[citations.length + 1]; 8 9 for (int i = 0; i citations.length...; i++) { 10 /* 11 if (citations[i] > citations.length) { 12 counts[citations.length...] += 1; 13 } else { 14 counts[citations[i]] += Math.min(1, citations[i]); // this...17 */ 18 counts[Math.min(citations[i], citations.length)]++; 19 20 } 21 22

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LeetCode 0274 - H-Index

each, and the other N − h papers have no more than h citations each.”...with no more than 3 citations each, her h-index is 3....Solution class Solution { public: int hIndex(vector& citations) { sort(citations.begin...(), citations.end()); for(int i = 0; i citations.size(); i++) { if(citations.size...() - i citations[i]) { return citations.size() - i; } }

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LeetCode 274. H-Index

排个序就好了 class Solution { public: int hIndex(vector& citations) { if(citations.size...()==0) return 0; sort(citations.begin(),citations.end()); int ans...=citations.size(); for(int i=0;icitations.size();i++) { if(citations

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Leetcode 题目解析之 H-Index

each, and the other N − h papers have no more than h citations each."...of them had received 3, 0, 6, 1, 5 citations respectively....3 citations each, his h-index is 3....) { if (citations == null) { return 0; } int h = 0, n = citations.length...; Arrays.sort(citations); for (int i = n - 1; i >= 0; i--) { if (citations

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Leetcode 274. H-Index

each of them had received 3, 0, 6, 1, 5 citations respectively....than 3 citations each, his h-index is 3....class Solution { public: int hIndex(vector& citations) { vector cnt(citations.size...() + 2, 0); for(int i = 0; i citations.size(); i++) { if(citations[i] >...citations.size()) cnt[citations.size()]++; else cnt[citations[i]]++; } for

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LeetCode 0275 - H-Index II

H-Index II Desicription Given an array of citations sorted in ascending order (each citation is a non-negative...each, and the other N − h papers have no more than h citations each.”...Example: Input: citations = [0,1,3,5,6] Output: 3 Explanation: [0,1,3,5,6] means the researcher has...with no more than 3 citations each, her h-index is 3....mid = (left + right) >> 1; if(citations[mid] >= citations.size() - mid) {

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016— H指数【LeetCode274】

H 指数 - 力扣（LeetCode）给你一个整数数组 citations ，其中 citations[i] 表示研究者的第 i 篇论文被引用的次数。计算并返回该研究者的 h 指数。...示例 2：输入：citations = [1,3,1] 输出：1 提示： n == citations.length 1 <= n <= 5000 0 citations[i] <= 1000...解题解法一思路首先数组进行排序，然后从第一个开始遍历，要是citations[i] citations.length - i，就更新max，遍历结束，返回max即可完成。...解决 class Solution { public int hIndex(int[] citations) { Arrays.sort(citations);...); int res=0; for(int i=0;i=citations.length-i){ res=citations.length-i

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H 指数

给你一个整数数组 citations ，其中 citations[i] 表示研究者的第 i 篇论文被引用的次数。计算并返回该研究者的 h 指数。...示例 1：输入：citations = [3,0,6,1,5] 输出：3 解释：给定数组表示研究者总共有5 篇论文，每篇论文相应的被引用了 3, 0, 6, 1, 5 次。 ...由于研究者有 3 篇论文每篇至少被引用了 3 次，其余两篇论文每篇被引用不多于 3 次，所以她的 h 指数是 3 示例 2：输入：citations = [1,3,1] 输出：1 方法一：排序...class Solution { public: int hIndex(vector& citations) { sort(citations.begin(), citations.end...()); int h = 0, i = citations.size() - 1; while (i >= 0 && citations[i] > h) {

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Leetcode 275. H-Index II

Follow up for H-Index: What if the citations array is sorted in ascending order?...class Solution { public: int hIndex(vector& citations) { int mid = -1, l = 0, r = citations.size...()-1; while(l <= r) { mid = (l + r) >> 1; if(citations.size()...- mid == citations[mid]) return citations[mid]; else if(citations.size() - mid > citations...[mid]) l = mid + 1; else r = mid - 1; } return citations.size() - r - 1;

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Leetcode 274. H-Index

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力扣经典150题第十一题：H指数

问题描述给你一个整数数组 citations ，其中 citations[i] 表示研究者的第 i 篇论文被引用的次数。计算并返回该研究者的 h 指数。...示例 2：输入：citations = [1,3,1] 输出：1 提示： n == citations.length 1 <= n <= 5000 0 citations[i] <= 1000...示例 2：输入：citations = [1, 3, 1] 输出：1 解决方案排序 + 线性扫描我们可以通过排序数组 citations 并进行线性扫描来解决该问题：首先对数组 citations...对于排序后的数组中第 i 个位置的值 citations[i]，表示至少有 i+1 篇论文被引用了 citations[i] 次。找到最大的 h 指数，使得至少有 h 篇论文被引用了 h 次。...import java.util.Arrays; public int hIndex(int[] citations) { Arrays.sort(citations); int n

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Leetcode 275. H-Index II

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LeetCode 275. H-Index II

题目现在变了，数列是拍好序的，题目要求对数效率，因为x只可能有一个那就二分咯 class Solution { public: int hIndex(vector& citations...) { if(citations.size()==0) return 0; int len = citations.size...; while(l<=r) { int mid = (l+r)/2; if(citations...[mid]>len-mid) { r = mid-1; } else if(citations[mid...else { return len-mid; } } if(lcitations

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LeetCode 275. H指数 II（二分查找）

进阶：本题中的 citations 数组是保证有序的。你可以优化你的算法到对数时间复杂度吗？...int i = citations.size()-1, papers = 0; while(i >= 0 && papers citations[i]) {...二分查找 class Solution { public: int hIndex(vector& citations) { int l = 0, r = citations.size...()-1, mid, n = citations.size(); //找第一个citations[i] >= n-i while(l <= r) {...mid = l +((r-l)>>1); if(citations[mid] < n-mid) l = mid+1; else if(citations

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数字问题-LeetCode 263、264、274、275

示例: 输入: citations = [3,0,6,1,5] 输出: 3 解题思路： ? 题目有点难以理解，可以直接看图，如上图所示，绿色正方形的边长即为h的值。...因此，需要对引用次数进行排序处理，然后寻找citations[i]>=i+1的位置即可！...class Solution { public: int hIndex(vector& citations) { sort(citations.begin(), citations.end...示例: 输入: citations = [0,1,3,5,6] 输出: 3 解题思路：使用274题的方法也可以解决这个问题，但在本题中给定了引用数是有序的！...class Solution { public: int hIndex(vector& citations) { int total = citations.size(

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Leetcode 题目解析之 H-Index II

Follow up for H-Index: What if the citations array is sorted in ascending order?...二分查找 public int hIndex(int[] citations) { int n = citations.length; int low = 0,...n - 1; while (low <= high) { int mid = low + (high - low) / 2; if (citations...[mid] == n - mid) { return n - mid; } else if (citations[mid] < n - mid)

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葵花宝典之机器学习：全网最重要的AI资源都在这里了（大牛，研究机构，视频，博客，书籍，Quora......）

Google Scholar： https://scholar.google.com/citations?...lang=en Google Scholar： https://scholar.google.com/citations?...lang=en Google Scholar： https://scholar.google.com/citations?..._Jordan Google Scholar： https://scholar.google.com/citations?...user=XF6Yk98AAAAJ&hl=en Quora： https://scholar.google.com/citations?

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如何搜索同时引用了两篇特定文章的论文？

1906.08237', timeout=2) paper1.keys() dict_keys(['abstract', 'arxivId', 'authors', 'citationVelocity', 'citations...Autoregressive Pretraining for Language Understanding' paper1["influentialCitationCount"] 462 len(paper1["citations..."]) 2616 paper1["citations"][0] {'arxivId': None, 'authors': [{'authorId': '1519065273', 'name': 'Jia-ying...co_citations = list(set(x['title'] for x in paper1['citations'] if x['isInfluential']) & set(x['title...'] for x in paper2['citations'] if x['isInfluential'])) len(co_citations) 158 print("\n".join(co_citations

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开发者如何利用Perplexity的Sonar大型语言模型

import requests def get_results_citations(api_key, query): """Retrieves citations from the Perplexity...Returns: A list of citations (URLs). """ url = "https://api.perplexity.ai/chat/completions...response = get_results_citations(api_key, query) # Extract content and citations content = response[...'choices'][0]['message']['content'] citations = response['citations'] # Print the results print("Content...:\n", content) print("\nCitations:") for i, citation in enumerate(citations, start=1): print(f"[{

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LeetCode 274. H指数（排序，哈希）

Baozi Training Leetcode solution 274: H-Index

LeetCode 0274 - H-Index

LeetCode 274. H-Index

Leetcode 题目解析之 H-Index

Leetcode 274. H-Index

LeetCode 0275 - H-Index II

016— H指数【LeetCode274】

H 指数

Leetcode 275. H-Index II

Leetcode 274. H-Index

力扣经典150题第十一题：H指数

Leetcode 275. H-Index II

LeetCode 275. H-Index II

LeetCode 275. H指数 II（二分查找）

数字问题-LeetCode 263、264、274、275

Leetcode 题目解析之 H-Index II

葵花宝典之机器学习：全网最重要的AI资源都在这里了（大牛，研究机构，视频，博客，书籍，Quora......）

如何搜索同时引用了两篇特定文章的论文？

开发者如何利用Perplexity的Sonar大型语言模型

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