原题:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
思路:关键是溢出的问题,32位数表示的范围为:-2^31—2^31-1,INT_MAX=2^31-1,INT_MIN=-INT_MAX-1,关于表示范围的疑问,可参考博客:https://www.sigmainfy.com/blog/2s-complement-int-max-int-min-difference.html,采用long long整形来表示数以防溢出报错。注意:正负数不必区分,可以自己找几个数试验一下。
class Solution {
public:
int reverse(int x){
long long res=0;
while(x!=0){
res=10*res+x%10;
x=x/10;
}
return (res>INT_MAX || res<INT_MIN)? 0:res;
}
};