翻转数值

Posted 2022-09-20 jest549

输入1234  输出4321

输入-1234 输出-4321

要求实现判断:输入数为-2^(32-1)~2^(32-1)-1,翻转数-2^(32-1)~2^(32-1)-1,在此范围输出翻转数,否则提示溢出.

实现:法一取余放入队列,取出队首乘倍数实现翻转

#include<iostream>

#include<queue>

using namespace std;

int dealFuntion(long long input);

long long mypow(int n);

int main(int argc, char* argv[])

    

    long long a=0;

    cout<<"输入需要翻转的数值:"<<endl;

    cin>>a;

    int b=dealFuntion(a);

    cout<<"翻转的数值为:"<<endl;

    cout<<b<<endl;

    return 0;

int dealFuntion(long long input)

if(input>2147483647||input<-2147483648)

cout<<"输入数溢出"<<endl;

return 0;

    queue<int> s1;

    long long i=0;

    int j=0;

    int symbol=1;

    int count=0;

    long long res;

    long long tmp;

    if(input ==0)

    

        return 0;

    

tmp=input;

    if(input<0)

    

        symbol=-1;

tmp=-input;

    

    if(tmp%10!=0)

    

        for(i=1;(j=(tmp/i)%10)!=0;i=i*10)

        

            s1.push(j);

            count++;

        

    

    else

       

        s1.push(0);

        count++;

        for(i=10;(j=(tmp/i)%10)!=0;i=i*10)

        

            s1.push(j);

            count++;

        

    

    for(;(!s1.empty());s1.pop())

    

        int tmp=s1.front();

        res=res+tmp*mypow(count-1);

if(res>2147483647)

            cout<<"翻转数溢出"<<endl;

return 0;

        count--;

    

return (int)res*symbol;

long long mypow(int n)

int i=0;

long long res=1;

for(i=1;i<=n;i++)

res=10*res;

return res;

 

 

法二:数值转为字符串,翻转字符串,字符串转为数值实现翻转

#include<iostream>

#include<sstream>

#include<algorithm>

#include<string>

#include<stdlib.h>

using namespace std;

string lltostring(long long input);

long long rev_stringtoll(string my_string);

void rev_result(long long input);

int main(int argc, char* argv[])

long long input;

cout<<"输入翻转数值"<<endl;

cin>>input;

rev_result(input);

return 0;

void rev_result(long long input)

long long my_ll;

int symbol=1;

if(input>2147483647||input<-2147483648)

cout<<"输入数溢出"<<endl;

return;

if(input==0)

cout<<"翻转结果为:"<<endl<<"0"<<endl;

return;

if(input<0)

symbol=-1;

input=-input;

my_ll=rev_stringtoll(lltostring(input))*symbol;

if(my_ll>2147483647||my_ll<-2147483648)

cout<<"翻转数溢出"<<endl;

else

cout<<"翻转结果为:"<<endl<<my_ll<<endl;

long long rev_stringtoll(string my_string)

reverse(my_string.begin(),my_string.end());

return strtoll(my_string.c_str(),NULL,10);

string lltostring(long long input)

stringstream my_stream;

string res_string;

my_stream<<input;

my_stream>>res_string;

return res_string;

/*

//实现字符串翻转

char *reverse_str(char *str)

if(NULL == str) //字符串为空直接返回

return str;

char *begin;

char *end;

begin = end = str;

while(*end != ‘\0‘) //end指向字符串的末尾

end++;

--end;

char temp;

while(begin < end) //交换两个字符

temp = *begin;

*begin = *end;

*end = temp;

begin++;

end--;

return str; //返回结果

void main()

char str[] = "123456";

printf(reverse_str(str));

*/