如何在java中对整数的数字求和？
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How to sum digits of an integer in java?
提问by Shane Larsen
I am having a hard time figuring out the solution to this problem. I am trying to develop a program in Java that takes a number, such as 321, and finds the sum of digits, in this case 3 + 2 + 1 = 6. I need all the digits of any three digit number to add them together, and store that value using the % remainder symbol. This has been confusing me and I would appreciate anyones ideas.
我很难找出这个问题的解决方案。我正在尝试用 Java 开发一个程序，它需要一个数字，例如 321，并找到数字的总和，在本例中为 3 + 2 + 1 = 6。我需要任何三位数的所有数字将它们加在一起，并使用 % 余数符号存储该值。这让我很困惑，我会很感激任何人的想法。
采纳答案by Ankur Singhal
public static void main(String[] args) {
int num = 321;
int sum = 0;
while (num > 0) {
sum = sum + num % 10;
num = num / 10;
}
System.out.println(sum);
}
Output
输出
6
回答by Simon Woo
The following method will do the task:
以下方法将完成任务：
public static int sumOfDigits(int n) {
String digits = new Integer(n).toString();
int sum = 0;
for (char c: digits.toCharArray())
sum += c - '0';
return sum;
}
You can use it like this:
你可以这样使用它：
System.out.printf("Sum of digits = %d%n", sumOfDigits(321));
回答by Arunkumar Papena
Here is a simple program for sum of digits of the number 321.
这是一个简单的程序，用于计算数字 321 的数字总和。
import java.math.*;
class SumOfDigits {
public static void main(String args[]) throws Exception {
int sum = 0;
int i = 321;
sum = (i % 10) + (i / 10);
if (sum > 9) {
int n = (sum % 10) + (sum / 10);
System.out.print("Sum of digits of " + i + " is " + n);
}else{
System.out.print("Sum of digits of " + i + " is " + sum );
}
}
}
Output:
Sum of digits of 321 is 6
Or simple you can use this..check below program.
或者简单的你可以使用这个..检查下面的程序。
public class SumOfDigits {
public static void main(String[] args)
{
long num = 321;
/* int rem,sum=0;
while(num!=0)
{
rem = num%10;
sum = sum+rem;
num=num/10;
}
System.out.println(sum);
*/
if(num!=0)
{
long sum = ((num%9==0) ? 9 : num%9);
System.out.println(sum);
}
}
回答by Ramesh Babu
Click here to see full program
Sample code:
示例代码：
public static void main(String args[]) {
int number = 333;
int sum = 0;
int num = number;
while (num > 0) {
int lastDigit = num % 10;
sum += lastDigit;
num /= 10;
}
System.out.println("Sum of digits : "+sum);
}
回答by krmanish007
In Java 8, this is possible in a single line of code as follows:
在 Java 8 中，这可以在一行代码中实现，如下所示：
int sum = Pattern.compile("")
.splitAsStream(factorialNumber.toString())
.mapToInt(Integer::valueOf)
.sum();
回答by CyprUS
It might be too late, but I see that many solutions posted here use O(n^2) time complexity, this is okay for small inputs, but as you go ahead with large inputs, you might want to reduce time complexity. Here is something which I worked on to do the same in linear time complexity.
可能为时已晚，但我看到这里发布的许多解决方案使用 O(n^2) 时间复杂度，这对于小输入是可以的，但是当您继续处理大输入时，您可能希望降低时间复杂度。这是我在线性时间复杂度上做同样的事情。
NOTE : The second solution posted by Arunkumar is constant time complexity.
注意：Arunkumar 发布的第二个解决方案是恒定时间复杂度。
private int getDigits(int num) {
int sum =0;
while(num > 0) { //num consists of 2 digits max, hence O(1) operation
sum = sum + num % 10;
num = num / 10;
}
return sum;
}
public int addDigits(int N) {
int temp1=0, temp2= 0;
while(N > 0) {
temp1= N % 10;
temp2= temp1 + temp2;
temp2= getDigits(temp2); // this is O(1) operation
N = N/ 10;
}
return temp2;
}
Please ignore my variable naming convention, I know it is not ideal. Let me explain the code with sample input , e.g. "12345". Output must be 6, in a single traversal.
请忽略我的变量命名约定，我知道这并不理想。让我用示例输入解释代码，例如“12345”。在单次遍历中输出必须为 6。
Basically what I am doing is that I go from LSB to MSB , and add digits of the sum found, in every iteration.
基本上我所做的是从 LSB 到 MSB ，并在每次迭代中添加找到的总和的数字。
The values look like this
这些值看起来像这样
Initially temp1 = temp2 = 0
最初 temp1 = temp2 = 0
N | temp1 ( N % 10) | temp2 ( temp1 + temp2 )
12345 | 5 | 5
1234 | 4 | 5 + 4 = 9 ( getDigits(9) = 9)
123 | 3 | 9 + 3 = 12 = 3 (getDigits(12) =3 )
12 | 2 | 3 + 2 = 5 (getDigits(5) = 5)
1 | 1 | 5 + 1 = 6 (getDigits(6) = 6 )
Answer is 6, and we avoided one extra loop. I hope it helps.
答案是 6，我们避免了一个额外的循环。我希望它有帮助。
回答by user1139428
May be little late ..but here is how you can do it recursively
可能有点晚了..但这里是你如何递归地做到这一点
public int sumAllDigits(int number) {
int sum = number % 10;
if(number/10 < 10){
return sum + number/10;
}else{
return sum + sumAllDigits(number/10);
}
回答by SlwoJams
Mine is more simple than the others hopefully you can understand this if you are a some what new programmer like myself.
我的比其他的更简单，希望你能理解这一点，如果你是一个像我这样的新程序员。
import java.util.Scanner;
import java.lang.Math;
public class DigitsSum {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int digit = 0;
System.out.print("Please enter a positive integer: ");
digit = in.nextInt();
int D1 = 0;
int D2 = 0;
int D3 = 0;
int G2 = 0;
D1 = digit / 100;
D2 = digit % 100;
G2 = D2 / 10;
D3 = digit % 10;
System.out.println(D3 + G2 + D1);
}
}
回答by Ridham Tarpara
This should be working fine for any number of digits and it will return individual digit's sum
这应该适用于任意数量的数字，它会返回单个数字的总和
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("enter a string");
String numbers = input.nextLine(); //String would be 55
int sum = 0;
for (char c : numbers.toCharArray()) {
sum += c - '0';
}
System.out.println(sum); //the answer is 10
}
回答by Chris Johnson
shouldn't you be able to do it recursively something like so? I'm kinda new to programming but I traced this out and I think it works.
你不应该像这样递归地做吗？我对编程有点陌生，但我追踪到了这一点，我认为它有效。
int sum(int n){
return n%10 + sum(n/10);
}
