[英]reversing an integer in java without a loop
This is an homework problem这是一道作业题
Is there a way tor reverse a number in Java without using any loops?有没有办法在不使用任何循环的情况下反转 Java 中的数字? The only solution I can think of is reversing it using String and then casting it back to an integer.
我能想到的唯一解决方案是使用 String 反转它,然后将其转换回 integer。
If you want to reverse a number withour using any loop you can use Recursion method call.
如果您想在不使用任何循环的情况下反转数字,您可以使用递归方法调用。 Following program is doing same
以下程序正在做同样的事情
public static void reverseMethod(int number) {
if (number < 10) {
System.out.println(number);
return;
} else {
System.out.print(number % 10);
reverseMethod(number / 10);
}
}
public static void main(String args[]) {
int num = 4567;
reverseMethod(num);
}
Even if you were to reverse the number by casting it into a String, you would still need a loop if you want the program to work when having ints of different sizes.即使您通过将其转换为字符串来反转数字,如果您希望程序在具有不同大小的整数时工作,您仍然需要一个循环。 If I were to make a method to reverse a number but could not do it with loops, I would probably do it with recursion (which still uses loops indirectly).
如果我要创建一个方法来反转一个数字但不能用循环来做,我可能会用递归来做(它仍然间接使用循环)。 The code will look something like this:
代码如下所示:
class Main {
public static void main(String[] args) {
String input = "1234"; // or scanner to take in input can be implemented
System.out.println(Integer.parseInt(reverseInt(input)));
}
public static String reverseInt(String x) {
if (x.length() == 1) {
return x;
} else {
return x.substring(x.length() - 1) + reverseInt(x.substring(0, x.length() - 1));
}
}
}
Hope this helps!希望这可以帮助!
By using reverse()
of StringBuilder
:通过使用
StringBuilder
reverse()
:
int number = 1234;
String str = String.valueOf(number);
StringBuilder builder = new StringBuilder(str);
builder.reverse();
number = Integer.parseInt(builder.toString());
System.out.println(number);
will print:将打印:
4321
if you want reverse method without loop and recursion then use this code如果您想要没有循环和递归的反向方法,请使用此代码
int a=12345;
int b,c,d,e,f;
b=a%10;
c=a%100/10;
d=a%1000/100;
e=a%10000/1000;
f=a%100000/10000;
System.out.println(b+","+c+","+d+","+e+","+f);
you can go like :你可以这样:
public int reverse(int x) {
String o = "";
if (x < 0) {
x *= -1;
String s = Integer.toString(x);
o += "-";
o += new StringBuilder(s).reverse().toString();
}
else {
String s = Integer.toString(x);
o += new StringBuilder(s).reverse().toString();
}
try {
int out = Integer.parseInt(o);
//System.out.println(s);
return out; }
catch (NumberFormatException e) {
return 0;
}
}
public class Tester{公共课测试员{
public static int findReverse(int num, int temp){
if(num==0){
return temp;
}else if(num<10){
return temp*10 + num; //up to this is stopping condition
}else{
temp = temp*10 + num%10;
return findReverse(num/10, temp);
}
}
public static void main(String args[]){
int num = 120021;
int reverseNum = findReverse(num, 0);
System.out.println(reverseNum);
if(num == reverseNum)
System.out.println(num +" is a palindrome!");
else
System.out.println(num +" is not a palindrome!");
}
} }
This will be fast.这会很快。
static int reverseNum(int num) {
StringBuilder sb = new StringBuilder(String.valueOf(num));
sb.reverse();
return Integer.parseInt(sb.toString());
}
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