Two, integer inversion
Give you a 32-bit signed integer x, and return the result of the inversion of each digit in x.
If the integer exceeds the 32-bit signed integer range [−231, 231 − 1] after the inversion, 0 is returned.
Assume that the environment does not allow storage of 64-bit integers (signed or unsigned).
1. My
class Solution {
public int reverse(int x) {
boolean check=false;
if(x<0){
x=-x;
check=true;
}
int minnum=-2147483648;
int maxnum=2147483647;
long y=0;
while(x!=0){
y=y*10+x%10;
x=x/10;
}
if(y<=minnum || y>=maxnum){
return 0;
}
if(check){
y=-y;
}
int m=(int)y;
return m;
}
}
2. Official
class Solution {
public int reverse(int x) {
int rev = 0;
while (x != 0) {
int pop = x % 10;
x /= 10;
if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
rev = rev * 10 + pop;
}
return rev;
}
}
作者：LeetCode
链接：https://leetcode-cn.com/problems/reverse-integer/solution/zheng-shu-fan-zhuan-by-leetcode/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
3. I saw a good one in the comment area
public int reverse(int x) {
int ans = 0;
while (x != 0) {
if ((ans * 10) / 10 != ans) {
ans = 0;
break;
}
ans = ans * 10 + x % 10;
x = x / 10;
}
return ans;
}