Given a 32-bit signed integer, reverse digits of an integer.
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.
解法一:
class Solution {
public:
int reverse(int x) {
int res = ;
while (x != ) {
if (abs(res) > INT_MAX / ) return ;
res = res * + x % ;
x /= ;
}
return res;
}
};
在贴出答案的同时,OJ 还提了一个问题 To check for overflow/underflow, we could check if ret > 214748364 or ret < –214748364 before multiplying by 10. On the other hand, we do not need to check if ret == 214748364, why? (214748364 即为 INT_MAX / )
为什么不用 check 是否等于 214748364 呢,因为输入的x也是一个整型数,所以x的范围也应该在 -2147483648~2147483647 之间,那么x的第一位只能是1或者2,翻转之后 res 的最后一位只能是1或2,所以 res 只能是 2147483641 或 2147483642 都在 int 的范围内。但是它们对应的x为 1463847412 和 2463847412,后者超出了数值范围。所以当过程中 res 等于 214748364 时, 输入的x只能为 1463847412, 翻转后的结果为 2147483641,都在正确的范围内,所以不用 check。
我们也可以用 long 型变量保存计算结果,最后返回的时候判断是否在 int 返回内,但其实题目中说了只能存整型的变量,所以这种方法就只能当个思路扩展了,参见代码如下:
解法二:
class Solution {
public:
int reverse(int x) {
long res = ;
while (x != ) {
res = * res + x % ;
x /= ;
}
return (res > INT_MAX || res < INT_MIN) ? : res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/7
类似题目:
参考资料:
https://leetcode.com/problems/reverse-integer/
https://leetcode.com/problems/reverse-integer/discuss/4060/My-accepted-15-lines-of-code-for-Java
https://leetcode.com/problems/reverse-integer/discuss/4056/Very-Short-(7-lines)-and-Elegant-Solution