Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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.
To check for overflow/underflow, we could check if ret > 214748364 or ret < –214748364 before multiplying by 10. On the other hand, we do not need to check if ret == 214748364, why?
Average Rating: 3.5 (360 votes)
public class Solution {
public int reverse(int x) {
int flag=x>0?1:-1,res=0;
x=x>0?x:-x;
while(x>0){
if(res*10.0 + x%10 > 2147483647)return 0;
res = res*10+x%10;
x/=10;
}
return res*flag;
}
}int reverse(int x) {
int flag=x>0?1:-1,res=0;
x=x>0?x:-x;
while(x>0){
if((2147483647.0-x%10)/10<res)return 0;
res=res*10+x%10;
x=x/10;
}
return res*flag;
}class Solution {
public:
int reverse(int x) {
int flag=x>0?1:-1,res=0;
x=x>0?x:-x;
while(x>0){
if(res*10.0+x%10 > 2147483647)return 0;
res = res*10+x%10;
x/=10;
}
return res*flag;
}
};class Solution:
# @param {integer} x
# @return {integer}
def reverse(self, x):
flag=1 if x>0 else -1
x=x if x>0 else -x
res=0
while x>0:
if res*10.0 + x%10 > 2147483647:return 0
res = res*10+x%10
x/=10
return res*flagLeetCode 7 Reverse Integer(C,C++,Java,Python)
原文地址：http://blog.csdn.net/runningtortoises/article/details/45556033
