Reverse Integer

题目描述：

https://leetcode-cn.com/problems/reverse-integer/

思路：从原数中分离出各个位的数值，并且倒置即可。

#include<cstdio>
#include<iostream>
#include<math.h>
#include<stdlib.h>
using namespace std;

int reverse(int x) 
{
	//只有一位则直接返回该值 
	if(x<10&&x>-10)
	    return x;
	int num = abs(x);
        long result = 0;
	//得到x是几位数 
	int length = 1; 
	while(((int)(num/pow(10,length)))!=0)
		length += 1;
	cout<<"length:"<<length<<endl;
	//翻转x
	while(length!=0)
	{
	    int a = num%10;
	    result += a*pow(10,length-1);
	    length -= 1;
	    num /= 10; 
	}
	result = x<0?((-1)*result):result;
	//result超出界限则返回0 
	if(result>2147483647||result<-2147483648) 
	    result = 0;	
	return result;   
}
    
int main(){
	int x = -321;
	int result = reverse(x);
	printf("%d\n",result);
	return 0;
}