Java 如何编写递归方法来返回整数中的数字总和？
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How to write a recursive method to return the sum of digits in an int?
提问by user1268088
So this is my code so far.
所以这是我到目前为止的代码。
public int getsum (int n){
int num = 23456;
int total = 0;
while (num != 0) {
total += num % 10;
num /= 10;
}
}
The problem is that i cant/know how to change this into a recursive method Im kind of new with recursion and i need some help on implementing this method to change it so its recursive.
问题是我不能/不知道如何将其更改为递归方法我对递归有点陌生，我需要一些帮助来实现此方法以将其更改为递归。
回答by Rahul Borkar
Here it is,
这里是，
//sumDigits function
int sumDigits(int n, int sum) {
// Basic Case to stop the recursion
if (n== 0) {
return sum;
} else {
sum = sum + n % 10; //recursive variable to keep the digits sum
n= n/10;
return sumDigits(n, sum); //returning sum to print it.
}
}
An example of the function in action:
运行中的函数示例：
public static void main(String[] args) {
int sum = sumDigits(121212, 0);
System.out.println(sum);
}
回答by me_digvijay
Try this:
尝试这个：
int getSum(int num)
{
total = total + num % 10;
num = num/10;
if(num == 0)
{
return total;
} else {
return getSum(num);
}
}
回答by Naytzyrhc
Short, recursive and does the job:
简短，递归并完成工作：
int getsum(int n) {
return n == 0 ? 0 : n % 10 + getsum(n/10);
}
回答by Silver
int getSum(int N)
{
int totalN = 0;
totalN += (N% 10);
N/= 10;
if(N == 0)
return totalN;
else
return getSum(N) + totalN;
}
回答by JerryFZhang
public static int digitSum (int n)
{
int r = n%10; //remainder, last digit of the number
int num = n/10; //the rest of the number without the last digit
if(num == 0)
{
return n;
} else {
return digitSum (num) + r;
}}
回答by TheArchon
This works for positive numbers.
这适用于正数。
public int sumDigits(int n) {
int sum = 0;
if(n == 0){
return 0;
}
sum += n % 10; //add the sum
n /= 10; //keep cutting
return sum + sumDigits(n); //append sum to recursive call
}
回答by nas
#include <iostream>
int useRecursion(int x);
using namespace std;
int main(){
int n;
cout<<"enter an integer: ";
cin>>n;
cout<<useRecursion(n)<<endl;
return 0;
}
int useRecursion(int x){
if(x/10 == 0)
return x;
else
return useRecursion(x/10) + useRecursion(x%10);
}
回答by Samuel
import java.util.Scanner;
public class Adder {
public static void main(String[] args){
Scanner input = new Scanner(System.in);
System.out.print("Enter a number: ");
System.out.println();
int number = input.nextInt();
System.out.println("The sum of the digits is " +adder(number));
}
public static int adder(int num){
int length = String.valueOf(num).length();
int first , last , sum;
if (length==1){
return num;
}
else
{
first = num /10;
last = num % 10;
sum = last + adder(first);
}
return sum;
}
}
回答by Dimuthu Maduranga
public static int sumOfDigit(int num){
int sum=0;
if (num == 0)
return sum;
sum = num%10 + sumOfDigit(num/10);
return sum;
}
public static void main(String args[]) {
Scanner input=new Scanner(System.in);
System.out.print("Input num : ");
int num=input.nextInt();
int s=sumOfDigit(num);
System.out.println("Sum = "+s);
}
}
}
回答by parsecer
I think it's the shortest so far. The input thing is up too you, though.
我认为这是迄今为止最短的。不过，输入也取决于你。
public static int getSum(int input) { //example: input=246
int sum=0;
if (input%10==input) { //246%10=6;
return input%10; //2%10=2
}
return input%10+getSum((input-input%10)/10); //(246-6)/10=24; 24%10=4
}
