Reverse Integer

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">Reverse Integer</span><span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);"> </span>

Total Accepted: 102425 Total Submissions: 436788 Difficulty: Easy

Reverse digits of an integer.

Example1: x = 123, return 321

Example2: x = -123, return -321

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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.

刚开始我的解法是：

class Solution {
public:
int reverse(int x) {
int result = 0;
int bei = 1;
int k;
int i = x>0?x:-x;
int j = x>0?x:-x;
while((i / 10) != 0)
{
bei = bei * 10;
i = i / 10;
}
while(j != 0)
{
k = j % 10;
result += k * bei;
bei = bei / 10;
j = j / 10;
}
if(x >= 0)
return result;
else
return -result;
}
};

这种解法想法比较不巧妙，同时也没有考虑溢出问题。

后来是百度出来的，这一点感觉很不好，稍微想不出来，就不自己想，就想百度，这点得改正。

class Solution {
public:
int reverse(int x) {
const int max = 0x7fffffff;  //int最大值
const int min = 0x80000000;  //int最小值
long long sum = 0;

while(x != 0)
{
int temp = x % 10;
sum = sum * 10 + temp;
if (sum > max || sum < min)   //溢出处理
{
sum = 0;
return sum;
}
x = x / 10;
}
return sum;
}
};