Reverse Integer
Question
- leetcode: Reverse Integer | LeetCode OJ
- lintcode: (413) Reverse Integer
Problem Statement
Reverse digits of an integer. Returns 0 when the reversed integer overflows (signed 32-bit integer).
Example
Given x = 123, return 321
Given 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.
Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
題解
如題目中所提示的，要考慮一下溢出的處理俯萎。
C++
class Solution {
public:
int reverse(int x) {
const int min = 0x80000000;
const int max = 0x7FFFFFFF;
long s = 0;
while(x != 0){
if (x >= min && x <= max){
int temp = x % 10;
s = s * 10 + temp;
x = (x - temp) / 10;
if(s < min || s > max){
s = 0;
break;
}
}
else{
s = 0;
break;
}
}
return s;
}
};
Error分析：
當(dāng)執(zhí)行到s = 964632435葫掉，此時(shí)還沒(méi)有溢出酣栈，如果繼續(xù)s * 10 + temp，此時(shí)s肯定會(huì)溢出唐含，因此會(huì)報(bào)錯(cuò)深浮。因此增加了一個(gè)判斷s是否溢出。
32位int范圍:-2147483648～2147483647掐场。
