reverse integer 之 c c++ python java

LeetCode OJ中的一道题Reverse Integer描述如下：

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321
Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

c:

int reverse(int x) {
if(x>2147483647){
return 0;
}
if(x<(-2147483648)){
return 0;
}
int flag=0;
long long ret=0;
while(x){
ret=ret*10+x%10;
x/=10;
}
if(ret>2147483647){
return 0;
}
if(ret<(-2147483648)){
return 0;
}
return ret;

}

结果5ms左右

c++

class Solution {
public:
int reverse(int x) {
if(x>2147483647){
return 0;
}
if(x<(-2147483648)){
return 0;
}
long long ret=0;
while(x){
ret=ret*10+x%10;
x/=10;
}
if(ret>2147483647){
return 0;
}
if(ret<(-2147483648)){
return 0;
}
return ret;
}
};

结果14ms左右

python：

class Solution:
# @return an integer
def reverse(self, x):
if x>2147483647:
return 0
if x<-2147483648:
return 0
flag=0
if x<0:
flag=1
x=-x
ret=0
while x!=0:
ret=ret*10+x%10
x=x/10
if flag==1:
ret=-ret
if ret>=-2147483648 and ret<=2147483647:
return ret
else:
return 0

结果63ms左右

java:

public class Solution {
public int reverse(int x) {
long ret=0l;
while(x!=0){
ret=ret*10+x%10;
x/=10;
}
if(ret>2147483647){
return 0;
}
if(ret<(-2147483648)){
return 0;
}
return (int)ret;
}
}

结果230ms左右
每次都是通过相同的1032次测试，四种不同的编程语言的运行时间差别还是挺大的。