kotlin中判断字符串_Kotlin程序计算字符串中每个字符的出现

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选择匿名的用户   2021-6-2 11:59   124   0

kotlin中判断字符串

Given a string, we have to count the occurrences of each character.

给定一个字符串,我们必须计算每个字符的出现次数。

Example:

例:

    Input:
    string = "Hello includehelp how are you !!"

    Output:
    { =5, a=1, !=2, c=1, d=1, e=4, H=1, h=2, i=1, l=4, n=1, o=3, p=1, r=1, u=2, w=1, y=1}

程序,用于计算Kotlin中字符串中每个字符的出现 (Program to count the occurrences of each character in a string in Kotlin)

In this program, we are using the concept of HashMap, which will contain the individual characters along with its occurrences in the string.

在此程序中,我们使用HashMap的概念,该概念将包含各个字符及其在字符串中的出现。

package com.includehelp.basic

import java.util.*

//Main Function, entry Point of Program
fun main(args: Array<String>) {

    //Input Stream
    val sc = Scanner(System.`in`)

    //input April20.string value
    println("Input String : ")
    val str: String = sc.nextLine()

    val characterHashMap: HashMap<Char, Int> = HashMap<Char, Int>()
    var c: Char
    // Scan string and build hash table
    for(i in str.indices){
        c = str[i]
        if (characterHashMap.containsKey(c)) {
            // increment count corresponding to c
            characterHashMap[c] = characterHashMap[c]!!+1

        } else {
            characterHashMap[c] = 1
        }
    }

    //print All Occurrence of character
    println("All character Count: $characterHashMap")
}

Output

输出量

Run 1:
Input String :
Hello includehelp how are you !!
All character Count: { =5, a=1, !=2, c=1, d=1, e=4, H=1, h=2, i=1, l=4, n=1, o=3, p=1, r=1, u=2, w=1, y=1}
---
Run 2:
Input String :
Hindustan India android Kotlin Java Flutter
All character Count: { =5, a=5, d=4, e=1, F=1, H=1, i=4, I=1, J=1, K=1, l=2, n=5, o=2, r=2, s=1, t=4, u=2, v=1}


翻译自: https://www.includehelp.com/kotlin/count-the-occurrences-of-each-character-in-a-string.aspx

kotlin中判断字符串

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