[LeetCode] 274. H-Index H指数
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."
Example:
Input:citations = [3,0,6,1,5]
Output: 3
Explanation:[3,0,6,1,5]means the researcher has5papers in total and each of them had
received3, 0, 6, 1, 5citations respectively.
Since the researcher has3papers with at least3citations each and the remaining
two with no more than3citations each, her h-index is3.
Note: If there are several possible values for h, the maximum one is taken as the h-index.
H指数(H index)是一个混合量化指标,可用于评估研究人员的学术产出数量与学术产出水平
可以按照如下方法确定某人的H指数:
将其发表的所有SCI论文按被引次数从高到低排序;
从前往后查找排序后的列表,直到某篇论文的序号大于该论文被引次数。所得序号减一即为H指数。
解法1: 先将数组排序,T:O(nlogn), S:O(1)。然后对于每个引用次数,比较大于该引用次数的文章,取引用次数和文章数的最小值,即 Math.min(citations.length-i, citations[i]),并更新 level,取最大值。排好序之后可以用二分查找进行遍历,这样速度会更快,可见:275. H-Index II H指数 II
解法2: Counting sort,T:O(n), S:O(n)。使用一个大小为 n+1 的数组count统计引用数,对于count[i]表示的是引用数为 i 的文章数量。从后往前遍历数组,当满足 count[i] >= i 时,i 就是 h 因子,返回即可,否则返回0。
为什么要从后面开始遍历? 为什么 count[i] >= i 时就返回?
一方面引用数引用数大于 i-1 的数量是i-1及之后的累加,必须从后往前遍历。另一方面,h 因子要求尽可能取最大值,而 h 因子最可能出现最大值的地方在后面,往前值只会越来越小,能尽快返回就尽快返回,所以一遇到 count[i] >= i 就返回。参考:Code_Granker
Java:
public class Solution {
public int hIndex(int[] citations) {
Arrays.sort(citations);
int level = 0;
for(int i = 0; i < citations.length; i++)
level = Math.max(level,Math.min(citations.length - i,citations[i]));
return level;
}
}
Java:
public class Solution {
public int hIndex(int[] citations) {
int n = citations.length;
int[] count = new int[n + 1];
for(int c : citations)
if(c >= n) count[n]++; //当引用数大于等于 n 时,都计入 count[n]中
else count[c]++;
for(int i = n; i > 0; i--) { //从后面开始遍历
if(count[i] >= i) return i;
count[i-1] += count[i]; //引用数大于 i-1 的数量是i-1及之后的累加
}
return 0;
}
}
Python: Counting sort.
class Solution(object):
def hIndex(self, citations):
"""
:type citations: List[int]
:rtype: int
"""
n = len(citations);
count = [0] * (n + 1)
for x in citations:
# Put all x >= n in the same bucket.
if x >= n:
count[n] += 1
else:
count[x] += 1 h = 0
for i in reversed(xrange(0, n + 1)):
h += count[i]
if h >= i:
return i
return h
Python: T: O(nlogn) O: O(1)
class Solution2(object):
def hIndex(self, citations):
"""
:type citations: List[int]
:rtype: int
"""
citations.sort(reverse=True)
h = 0
for x in citations:
if x >= h + 1:
h += 1
else:
break
return h
Python: T: O(nlogn) O: O(n)
class Solution3(object):
def hIndex(self, citations):
"""
:type citations: List[int]
:rtype: int
"""
return sum(x >= i + 1 for i, x in enumerate(sorted(citations, reverse=True)))
Python:
class Solution(object):
def hIndex(self, citations):
"""
:type citations: List[int]
:rtype: int
"""
if not citations: return 0
return max([min(i + 1, c) for i, c in enumerate(sorted(citations, reverse=True))])
C++:
class Solution {
public:
int hIndex(vector<int>& citations) {
sort(citations.begin(), citations.end(), greater<int>());
for (int i = 0; i < citations.size(); ++i) {
if (i >= citations[i]) return i;
}
return citations.size();
}
};
类似题目:
[LeetCode] 275. H-Index II H指数 II
All LeetCode Questions List 题目汇总
[LeetCode] 274. H-Index H指数的更多相关文章
- [LeetCode] 275. H-Index II H指数 II
Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize ...
- [LeetCode] H-Index II 求H指数之二
Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize ...
- leetcode@ [274/275] H-Index & H-Index II (Binary Search & Array)
https://leetcode.com/problems/h-index/ Given an array of citations (each citation is a non-negative ...
- LeetCode 274
H-Index Given an array of citations (each citation is a non-negative integer) of a researcher, write ...
- [LeetCode#274]H-Index
Problem: Given an array of citations (each citation is a non-negative integer) of a researcher, writ ...
- Sed命令n,N,d,D,p,P,h,H,g,G,x解析3
摘自:https://blog.csdn.net/WMSOK/article/details/78463199 Sed命令n,N,d,D,p,P,h,H,g,G,x解析 2017年11月06日 23: ...
- sed命令n,N,d,D,p,P,h,H,g,G,x解析2
摘自: https://blog.csdn.net/xiexingshishu/article/details/50514132 sed命令n,N,d,D,p,P,h,H,g,G,x解析 2016年0 ...
- oc 与 swift 之间的桥接文件 (ProjectNmae-Bridging-Header.h) (ProjectNmae-Swift.h)
oc 与 Swift 是2种不同的语言, oc代码只能写带oc文件里, Swift代码只能写在Swift文件里, 虽然2者不同语言, 但却能互相调用, 不过需要进行一下桥接, 就是下面的2个文件 (P ...
- 86-Money Flow Index 资金流量指数指标.(2015.7.3)
Money Flow Index 资金流量指数指标 计算: 1.典型价格(TP)=当日最高价.最低价与收盘价的算术平均值 2.货币流量(MF)=典型价格(TP)×N日内成交金额 3.如果当日MF> ...
随机推荐
- ViCANdo新版本发布(PART1) | 点云库(PCL)集成
激光雷达 随着智能驾驶技术的发展,激光雷达迅速的进入工程师的视野,不管是机械式.MEMS还是纯固态激光雷达,本质上都是以一定的速度扫描照射区域,在此过程中激光雷达不断的发出激光并接收反 ...
- 关于缩短cin时间的方法
std::ios::sync_with_stdio(false);
- Hikari java数据库连接池实战
环境InterlliJ2016.3 MySQL5.7.12 pom依赖: <dependency> <groupId>com.zaxxer</groupId> & ...
- NOI2019游记 —— 夏花般绚烂,繁星般璀璨
NOI 2019 游记 夏花般绚烂,繁星般璀璨 打算写成两个形式 Dairy Day -1 早早就到gzez集训了20几天,对整体的环境熟悉很多 在gzez看了场LNR Day 2 然后回到宾馆搞了个 ...
- mybatis配置打印sql
mybatis配置打印sql: <settings> <setting name="logImpl" value="STDOUT_LOGGING&quo ...
- Alpha冲刺(8/10)——追光的人
1.队友信息 队员学号 队员博客 221600219 小墨 https://www.cnblogs.com/hengyumo/ 221600240 真·大能猫 https://www.cnblogs. ...
- linux下安装Sublime Text3并将它的快捷方式放进启动器中和卸载Sublime
Sublime Text是一个代码编辑器,我主要是用它来编辑python.下面就来简单说明下它在linux的安装过程吧! 1.添加sublime text3的仓库 首先按下快捷键ctrl+alt+t打 ...
- go 学习 (一):环境配置
Go 下载地址:https://golang.google.cn/dl/ 右键我的电脑 --> 左上方 “高级系统设置” ---> 环境变量 --> 第二个菜单栏 “系统变 ...
- LeetCode 1093. Statistics from a Large Sample
原题链接在这里:https://leetcode.com/problems/statistics-from-a-large-sample/ 题目: We sampled integers betwee ...
- C++之Lambda研究
目录 目录 1 1. 前言 1 2. 示例1 1 3. 示例2 2 4. 示例3 3 5. 示例4 3 6. 示例5 6 7. 匿名类规则 6 8. 参考资料 7 1. 前言 本文代码测试环境为“GC ...