思路:
先判定符号,整型范围[-2^32,2^32]
取余除10操作,依次进行,越界返回0
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
#define IMAX numeric_limits<int>::max()
#define IMIN numeric_limits<int>::min()
class Solution {
public:
int reverse(int x) {
int sign= x>0?1:-1;
if(x==IMIN)return 0;
else x = abs(x);
int res=0,count=0;
for(;x;x/=10)
{
if(count>9)return (sign==1)?IMAX:IMIN;
res = res*10;
if(count==8){ //倒数第二位向后看会不会越界,越界返回0
if(sign==1&&res>IMAX/10) return 0;
if(sign==-1&&res*sign<IMIN/10) return 0;
}
if(count==9){
if(sign==1&&(IMAX-res<=x%10))return 0;//最后一位向后看不会越界,越界返回0
if(sign==-1&&(res*sign-IMIN<=x%10)) return 0;
}
count++;
res=x%10+res;
}
return res*sign;
}
};