public class Solution {
public int reverse(long a) {//题目中提到如果超过32-bit的返回0,想到的解决办法是用long来存,判断结果是否大于Integer.MAX_VALUE或小于Integer.MIN_VALUE
long temp = 0;
int res = 0;
if(a < 0){
a = Math.abs(a);
while(a != 0){
temp = temp * 10 + (a % 10);
a = a / 10;
}
if(-temp < java.lang.Integer.MIN_VALUE){
return 0;
}
else{
res = -(int)temp;
}
}
else{
while(a != 0){
temp = temp * 10 + (a % 10);
a = a / 10;
}
if(temp > java.lang.Integer.MAX_VALUE){
return 0;
}
else{
res = (int)temp;
}
}
return res;
}
}
leetcode [Reverse Integer]
最新推荐文章于 2020-01-31 13:51:08 发布