I want to read a number as a String, and split its characters to an integer array, and find the sum of it's digits by looping through that integers array.
我想读取一个数字作为字符串,并将其字符拆分为整数数组,并通过循环遍历整数数组找到它的数字之和。
This is my code so far:
到目前为止这是我的代码:
public static void main(String[] args) {
Scanner S = new Scanner(System.in);
String Number = S.next();
int counterEnd = Number.length();
int sum = 0 ;
for ( int i = 0 ; i < counterEnd ; i++) {
sum += sum + (Number.charAt(i));
}
System.out.println(sum);
}
Unfortunately, this code prints the sum of ASCII not the digits.
不幸的是,此代码打印ASCII的总和而不是数字。
4 个解决方案
#1
5
You can subtract the '0'
character (i.e. '1' - '0'
is 49 - 48
= 1
):
你可以减去'0'字符(即'1' - '0'是49 - 48 = 1):
sum += Number.charAt(i) - '0';
#2
4
You could, has @August suggested, substract the character '0'
to obtain the numeric value of the character (I find this approach kind of hackish). Or you can use Character.getNumericValue
to achieve this:
@August建议你可以减去字符'0'以获得字符的数值(我发现这种方法有点hackish)。或者您可以使用Character.getNumericValue来实现此目的:
sum += Character.getNumericValue(Number.charAt(i)); //note it's sum += theDigit; not sum += sum + theDigit
You might also want to look at the enhanced for loop, as you basically don't need the index here:
您可能还想查看增强的for循环,因为您基本上不需要索引:
for(char c : Number.toCharArray()) {
sum += Character.getNumericValue(c);
}
As of Java 8 you could also do it like this:
从Java 8开始,您也可以这样做:
int sum = Number.chars().map(Character::getNumericValue).sum();
It basically gets a Stream
of the characters in the String
, map each character to its corresponding numeric value and sum them.
它基本上获取String中字符的Stream,将每个字符映射到其对应的数值并对它们求和。
#3
0
sum+= Integer.parseInt(String.valueOf(Number.charAt(i)));
#4
0
You can do it like this:
你可以这样做:
public static void main(String[] args) {
Scanner S = new Scanner(System.in);
String n = S.next();
int sum = 0;
for (int i = 0; i < n.length(); i++) {
sum += Integer.parseInt(n.substring(i, i+1));
}
System.out.println(sum);
}
Note: Replacing the body of the for loop with:
注意:用以下代码替换for循环的主体:
int offset = (n.substring(i, i+1).equals("-")) ? 1 : 0;
sum += Integer.parseInt(n.substring(i, i+1+offset));
i+=offset;
Will allow the program to take negative numbers. Ex: Inputting
将允许该程序采取负数。例如:输入
-24
Would return a positive 2.
会回归正面2。
#1
5
You can subtract the '0'
character (i.e. '1' - '0'
is 49 - 48
= 1
):
你可以减去'0'字符(即'1' - '0'是49 - 48 = 1):
sum += Number.charAt(i) - '0';
#2
4
You could, has @August suggested, substract the character '0'
to obtain the numeric value of the character (I find this approach kind of hackish). Or you can use Character.getNumericValue
to achieve this:
@August建议你可以减去字符'0'以获得字符的数值(我发现这种方法有点hackish)。或者您可以使用Character.getNumericValue来实现此目的:
sum += Character.getNumericValue(Number.charAt(i)); //note it's sum += theDigit; not sum += sum + theDigit
You might also want to look at the enhanced for loop, as you basically don't need the index here:
您可能还想查看增强的for循环,因为您基本上不需要索引:
for(char c : Number.toCharArray()) {
sum += Character.getNumericValue(c);
}
As of Java 8 you could also do it like this:
从Java 8开始,您也可以这样做:
int sum = Number.chars().map(Character::getNumericValue).sum();
It basically gets a Stream
of the characters in the String
, map each character to its corresponding numeric value and sum them.
它基本上获取String中字符的Stream,将每个字符映射到其对应的数值并对它们求和。
#3
0
sum+= Integer.parseInt(String.valueOf(Number.charAt(i)));
#4
0
You can do it like this:
你可以这样做:
public static void main(String[] args) {
Scanner S = new Scanner(System.in);
String n = S.next();
int sum = 0;
for (int i = 0; i < n.length(); i++) {
sum += Integer.parseInt(n.substring(i, i+1));
}
System.out.println(sum);
}
Note: Replacing the body of the for loop with:
注意:用以下代码替换for循环的主体:
int offset = (n.substring(i, i+1).equals("-")) ? 1 : 0;
sum += Integer.parseInt(n.substring(i, i+1+offset));
i+=offset;
Will allow the program to take negative numbers. Ex: Inputting
将允许该程序采取负数。例如:输入
-24
Would return a positive 2.
会回归正面2。