Leetcode Reverse Integer 中的溢出处理

Question

原题有这样的提示“Didyounoticethatthereversedintegermightoverflow?Assumetheinputisa32-bitinteger,thenthereverseof1000000003overflows.Howshouldyouhandlesuchcases?”看到有人写了一个这样的答案intreverse(intx){intflag=x>0?1:-1,res=0;x=x>0?x:-x;while(x>0){if((2147483647.0-x%10)/10

Answer 1

(2147483647.0-x%10)/10
我的做法是直接用long型:
publicintreverse(intx){
intsign=1;
if(x<0){
sign=-1;
x*=-1;
}
longresult=0;
while(x!=0){
result=result*10+x%10;
if(sign*result>Integer.MAX_VALUE||
sign*resultreturn0;
x=x/10;
}
return(int)result*sign;
                            

Answer 2

它判断的是每轮的结果是否超过intif((2147483647.0-x%10)/10