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leetcode_7_Reverse Integer
描述:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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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.
Update (2014-11-10):
Test cases had been added to test the overflow behavior.
思路：
思路倒是简洁明了，将整数的非符号部分的各位存储到一个数组里面，然后将各位组装起来就成为一个新的整数。当然，溢出，符号都是需要考虑的。
代码：
int reverse(int x) {
int i=0;
int num[20]={0};
long sum=0;
int flag=0;
if(x<0)
{x=-x;flag=1;}
while(x!=0)
{
num[i++]=x%10;
x/=10;
}
for(int j=0;j<i;j++)
{
sum=sum*10+num[j];
}
if(flag)
sum=-sum;
return sum;
}结果：
leetcode_7_Reverse Integer