输入一个有符号整数，输出该整数的反转值。
[编程题]整数反转
先对绝对值反转，反转完成后再添加符号
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String strNum;
while((strNum = br.readLine()) != null){
int num = Integer.parseInt(strNum);
System.out.println(solve(num));
}
}
// 反转整数
private static int solve(int num) {
boolean flag = false;
// 记录数字的符号
if(num < 0) flag = true;
num = Math.abs(num);
int result = 0;
while(num > 0){
result = result * 10 + (num % 10);
num /= 10;
}
return flag ? -result: result;
}
}
发表于 2020-11-10 21:54:35
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int flag = 1; if (n < 0 ) { flag = -1; n = -n; }
StringBuilder sb = new StringBuilder(String.valueOf(n));
System.out.println(Integer.valueOf(sb.reverse().toString()) * flag);
return;
}
}
发表于 2019-02-13 20:31:59
回复(0)
#include <bits/stdc++.h>
usingnamespacestd;
intmain()
{
string s;
cin>>s;
if(s.length()==1&&s[0]=='0'){cout<<0;return0;}
intflag = 0;
inti = 0;
if(s[0]=='-'){
flag = 1;
}
reverse(s.begin(),s.end());
if(flag==0)
{
while(s[i]=='0')i++;
for(i;i<s.length();i++)
cout<<s[i];
}
if(flag==1){
while(s[i]=='0')i++;
cout<<'-';
for(i;i<s.length()-1;i++)
cout<<s[i];
}
return0;
}
编辑于 2018-09-12 12:59:16
回复(0)
#include<stdio.h>
#include<math.h>
int main()
{
int num;
scanf("%d", &num);
bool isnegative = false;
if (num < 0){
isnegative = true;
num = num * -1;
}
int t_num[4];
int i = 0;
while(num > 10){
t_num[i] = num % 10;
num = num / 10;
i++;
}
t_num[i] = num % 10;
num = 0;
for(int j=i; j>=0; j--){
num = num + t_num[i-j]*pow(10, j);
}
if(isnegative){
num = -1 * num;
}
printf("%d", num);
}
编辑于 2018-09-10 21:19:16
回复(0)
//0-1背包问题 //设m为背包剩余容量
//n为可能的物品集合
import java.util.*;
public class Main {
static int re =0;//输出结果
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m= scanner.nextInt();
int temp = n*(n+1)/2;
//判断[0-1]累加是否可以组成m
if(m>temp){
System.out.println(0);
return;
}else if(m==temp){
System.out.println(1);
}else{
tool(n,m);//递归查找所有可能的组合
}
System.out.println(re);
}
private static void tool(int n, int m) {
if (n > 0&&m>0) {//当m剩余数和n剩余数大于0时表示可以继续查找
if(m-n==0){//当背包剩余0时表示找到结果
re++;
}
//对任意一个物品n，都有是否放入背包的可能
tool(n-1,m);//不放
tool(n-1,m-n);//放
}
}
}
发表于 2018-09-07 23:08:15
回复(0)
import java.util.Scanner;
public class Main {
public static int reverse(int x) {
int rev = 0;
while (x != 0) {
rev = rev * 10 + +x % 10;
x /= 10;
}
return rev;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int input = in.nextInt();
System.out.println(reverse(input));
}
}
发表于 2018-09-15 07:41:28
回复(3)
// 简单几行代码，取余和取模的运算
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
int n = in.nextInt();
int res = 0;
while(n!=0){
res = res*10+n%10;
n /=10;
}
System.out.println(res);
}
}
发表于 2024-09-12 11:46:12
回复(0)
import java.util.*;
public class Main
{
public static int reverse(int x)
{
int rev = 0;
while (x != 0)
{
int pop = x % 10;
x /= 10;
if(rev > Integer.MAX_VALUE / 10 || (rev == Integer.MAX_VALUE && pop > 7))
{
return 0;
}
if(rev < Integer.MIN_VALUE / 10 || (rev == Integer.MIN_VALUE && pop < -8))
{
return 0;
}
rev = rev * 10 + pop;
}
return rev;
}
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
int input = in.nextInt();
System.out.println(reverse(input));
}
}
编辑于 2019-05-02 19:58:46
回复(0)
#include <iostream> using namespace std; long long myint; int main(){ cin>>myint; if(myint < 0) cout<<"-"; if(myint == 0) {cout<<0;return 0;} myint = abs(myint); int inx = 10; while(myint / inx != 0){ while (myint%10 == 0) { myint/=10; } cout<<myint%10; myint/=10; } if(myint!=0) cout<<myint; }
发表于 2025-03-31 23:33:53
回复(0)
public class Program {
public static void Main() {
string line;
while ((line = System.Console.ReadLine ()) != null) { // 注意 while 处理多个 case
string[] tokens = line.Split();
int input = int.Parse(tokens[0]);
int result = 0;
while(input != 0){
result = 10 * result + input%10;
input /= 10;
}
System.Console.WriteLine(result);
}
}
}
发表于 2024-05-09 08:51:52
回复(0)
#include <iostream> #include <climits> // 用于访问 INT_MAX 和 INT_MIN int reverse(int x) { long long reversed = 0; // 使用 long long 防止反转过程中溢出 while (x != 0) { reversed = reversed * 10 + x % 10; x /= 10; // 检查是否溢出 if (reversed > INT_MAX || reversed < INT_MIN) { return 0; // 溢出时返回0 } } return (int)reversed; } int main() { int number = 0; std::cout << "Enter an integer: "; std::cin >> number; int reversedNumber = reverse(number); std::cout << "Reversed integer: " << reversedNumber << std::endl; return 0; } </climits></iostream>
发表于 2024-05-07 15:01:50
回复(0)
#include<iostream>
int main()
{
int x;
while(std::cin>>x)
{
int num=0;
if(x>0)
{
while(x/10>0)
{
num=10*num+(x%10);
x/=10;
}
num=10*num+(x%10);
std::cout<<num<<std::endl;
}
if(x<0)
{
x=-x;
while(x/10>0)
{
num=10*num+(x%10);
x/=10;
}
num=-(10*num+(x%10));
std::cout<<num<<std::endl;
}
if(x==0)std::cout<<0<<std::endl;
}
return 0;
}
int main()
{
int x;
while(std::cin>>x)
{
int num=0;
if(x>0)
{
while(x/10>0)
{
num=10*num+(x%10);
x/=10;
}
num=10*num+(x%10);
std::cout<<num<<std::endl;
}
if(x<0)
{
x=-x;
while(x/10>0)
{
num=10*num+(x%10);
x/=10;
}
num=-(10*num+(x%10));
std::cout<<num<<std::endl;
}
if(x==0)std::cout<<0<<std::endl;
}
return 0;
}
发表于 2020-09-08 17:47:32
回复(0)
#include "stdlib.h"
#include "stdio.h"
int main()
{
int a,len,i;
char s[100]={0},d[100]={0};
while(scanf("%d",&a) != EOF)
{
sprintf(s,"%d",a);
len = strlen(s);
d[0] = '-';
if(s[0] == '-')
for(i = 0;i<len;i++)d[i+1] = s[len-i-1];
else
for(i = 0;i<len;i++)d[i] = s[len-i-1];
a = atoi(d);
printf("%d\n",a);
}
}
发表于 2020-01-25 02:31:46
回复(0)
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
int integer;
cin >> integer;
if(integer == 0)
{
cout << 0;
return 0;
}
int abs_value;
abs_value = (integer >= 0) ? integer : -integer;
string str_int;
str_int = to_string(abs_value);
reverse(str_int.begin(),str_int.end());
string reverse_value;
for(unsigned i = 0; i < str_int.size(); i++)
{
if(str_int[i] != '0')
{
reverse_value = str_int.substr(i);
break;
}
}
cout << ((integer >= 0) ? reverse_value : "-" + reverse_value);
return 0;
}
编辑于 2019-04-22 09:43:07
回复(0)
问题信息
通过挑战的用户
查看代码-
2022-11-22 16:18:46
-
2022-10-31 11:40:08
-
2022-10-27 15:34:02
-
2022-10-23 16:41:14 -
2022-10-10 18:30:43
相关试题
-
扫描二维码，关注牛客网
- 意见反馈
-
下载牛客APP，随时随地刷题
扫一扫，把题目装进口袋
- 求职之前，先上牛客
-
扫描二维码，进入QQ群
扫描二维码，关注牛客公众号
- 公司地址：北京市朝阳区北苑路北美国际商务中心K1座一层-北京牛客科技有限公司
- 联系方式：010-60728802 投诉举报电话：010-57596212（朝阳人力社保局）
- 牛客科技© All rights reserved admin@nowcoder.com
- 京ICP备14055008号-4 增值电信业务经营许可证 营业执照 人力资源服务许可证
-
京公网安备
11010502036488号
