完整的Java代码如下: importjava.util.Scanner;publicclassReverseString{publicstaticvoidmain(String[]args){Scannerscanner=newScanner(System.in);System.out.print("请输入一个字符串:");Stringinput=scanner.nextLine();scanner.close();StringBuilderstringBuilder=newStringBuilder();for(inti=0;i...
Java实现代码如下: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{publicintreverse(int x){long res=0;while(x!=0){res=res*10+x%10;x/=10;if(res>Integer.MAX_VALUE||res<Integer.MIN_VALUE)return0;}return(int)res;}}...
int变成string，string变成chararray，chararray倒序遍历变回string string变回int 记得判断正负。也可以用/10的余数取数字然后再乘10加回来 例如：public String reverseSting(String inputString) { char[] inputStringArray = inputString.toCharArray();String reverseString = "";for (int i = inputS...
On Crunchify, we have published more than 500 Java Tutorials and in this tutorial we will go over steps on how to reverse a string in Java? There are 7
Basic Java Program to Reverse anintArray In this first example, we take the size of array and the elements of array as input. We consider a functionreversewhich takes the array (here array) and the size of an array as the parameters. Inside the function, we initialize a new array. The...
import java.math.BigInteger; import java.util.Scanner; public class Test09 { public static void main(String[] args) { Scanner input = new Scanner(System.in); System.out.println("请输入一个整型数字:"); // 保存用户输入的数字 int num = input.nextInt(); ...
int[] array = {1, 2, 3}; Collections.reverse(Arrays.asList(array)); //Does not reverses the array 2. Swapping Array Elements in For Loop Another simple and straightforward way is to iterate through the array swapping them from the beginning position with the elements in the last position...
Integer.Reverse(Int32) 方法 参考 反馈 定义 命名空间: Java.Lang 程序集: Mono.Android.dll 返回通过反转指定 int 值的二进制补二进制表示形式中的位顺序获得的值。 C# 复制 [Android.Runtime.Register("reverse", "(I)I", "")] public static int Reverse (int i); 参数 i Int32 要撤消的...
Java: 可以AC的不严谨做法。 Time Complexity - O(lgx), Space Complexity - O(1) publicclassSolution {publicintreverse(intx) {intres = 0;while(x != 0) {if(res > Integer.MAX_VALUE / 10 || res < Integer.MIN_VALUE / 10) {return0; ...
classSolution {public:intreverse(intx) {intret =0;while(x !=0){//handle overfolow or underflowif(abs(ret) >214748364){return0; } ret= ret *10+ x %10; x/=10; }returnret; } }; Java: publicintreverse(intx) {intret = 0;while(x != 0) {//防止溢出,只要大于MAX/10就算溢出!