找出以d或e字母结尾的单词出现的次数

[编程题]找出以d或e字母结尾的单词出现的次数

热度指数：101 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

给出一组字符串，找出其中以小写的d或e字母结尾的单词出现的次数，结果以字典格式打印。其中不以d或e结尾的单词，不在结果统计中出现

输入描述:

若干个字符串,它们以逗号进行分隔，样例如下所示:

abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,hard,abc,head,tail,middle,head,today,tail,rain,end,cup,apple

输出描述:

以字典格式打印，样例输出如下所示:

{head=12, apple=6, middle=6, end=6, hard=1}

示例1

输入

abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,abc,head,tail,middle,head,today,tail,rain,end,cup,apple,hard,abc,head,tail,middle,head,today,tail,rain,end,cup,apple

输出

{head=12, apple=6, middle=6, end=6, hard=1}

算法知识视频讲解

Java

阿巴阿巴巴啦啦能量

用，分割，将空格字符串跳过

import java.util.*;

public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String ss = scanner.next();
count(ss);
}

public static void count(String ss){
String str[] = ss.split(",");
Map<String,Integer> map = new HashMap<>();
// 小写e或者d结尾单词出现次数
for(String s:str){
int len = s.length();
if(len==0){
continue;
}
if(s.lastIndexOf("e")==len-1 || s.lastIndexOf("d")==len-1){
map.put(s,map.getOrDefault(s,0)+1);
}
}
System.out.println(map);
}
}

发表于 2024-09-03 16:23:36 回复(0)