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
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)
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 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)
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)
#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)
/*************************************************************************
> File Name: xiangfan.c
> Author:
> Mail:
> Created Time: 2019年04月21日 星期日 04时56分55秒
************************************************************************/
#include<stdio.h>
int xiangfan(int num)
{
int ago = num;
int temp = 0;
while(ago)
{
temp = temp*10 + ago%10;
ago = ago/10;
}
return temp;
}
int main()
{
int num;
int result;
scanf("%d",&num);
result = xiangfan(num);
printf("%d\n",result);
return 0;
}
发表于 2019-04-21 20:04:19
回复(0)
#include <bits/stdc++.h>
using namespace std;
int main()
{
string data;
getline(cin,data);
reverse(data.begin(),data.end());
int len=data.length();
if(len==1&&data[0]=='0')
{
cout<<0;
return 0;
}
int index=0;
while(data[index]=='0')
{
++index;
}
if(data[len-1]=='-')
{
cout<<"-";
len=len-1;
}
for(int i=index;i<len;++i)
{
cout<<data[i];
}
return 0;
}
using namespace std;
int main()
{
string data;
getline(cin,data);
reverse(data.begin(),data.end());
int len=data.length();
if(len==1&&data[0]=='0')
{
cout<<0;
return 0;
}
int index=0;
while(data[index]=='0')
{
++index;
}
if(data[len-1]=='-')
{
cout<<"-";
len=len-1;
}
for(int i=index;i<len;++i)
{
cout<<data[i];
}
return 0;
}
/*
可能为大数问题，无法存储数字而溢出，使用string
闭关进行反转，随后确定反转后的字符串第一个不为零的索引，并确定是否为复数，进行输出
*/
发表于 2019-04-17 13:07:50
回复(0)
链接：https://www.nowcoder.com/questionTerminal/14733e0bfa9b474ba7cbe0bb2e459731 来源：牛客网#include<iostream>usingnamespacestd;intmain(){intin;cin >> in;intout = 0;while(in){out = out*10 + (in%10);in /= 10;}cout << out << endl;return0;}/*#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){int in;cin >> in;int flag = in>=0 ? 1 : -1;in = abs(in);int out = in;int bit = 1;while(out / 10){out /= 10;bit ++;}vector<int> v(bit);out = in;for(int i=0; i<bit; i++){v[bit-i-1] = out % 10;out /= 10;}reverse(v.begin(),v.end());for(int i=0; i<bit; i++){out = out*10 + v[i];}out *= flag;cout << out << endl;return 0;}*/
发表于 2019-04-10 23:16:41
回复(0)
public class Main {
public static void main(String[] args) throws IOException {
//把int类型转换成String类型的数据,再通过StringBuilder进行反转
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
String s = reader.readLine();
int num = Integer.parseInt(s);
int i = Math.abs(num);
if (num < 0) {
System.out.println(-1 * Integer.parseInt(new StringBuilder(String.valueOf(i)).reverse().toString()));
} else {
System.out.println(Integer.parseInt(new StringBuilder(String.valueOf(i)).reverse().toString()));
}
}
}
发表于 2019-04-09 14:59:27
回复(0)
问题信息
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