Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
public int reverse(int x) {
long res = 0; //根据note,使用long类型
while(x != 0) {
res = res * 10 + x % 10;
x /= 10;
if(res < Integer.MIN_VALUE || res > Integer.MAX_VALUE) {
return 0;
}
}
return (int)res;
}
}
Time Complexity = O(log(x)) 以10为底
Space Complexity = O(1)
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