problem
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: [$−2^{31}$, $2^{31} − 1$]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
analysis
- 因为反转可能出现溢出的情况，即超过32bit
int所能表示的最大（小）值，所以我习惯上先统一使用unsigned int来表示反转后的值，再去考虑溢出的情况； - leetcode-7与leetcode-8都有判断整数溢出的情况，俺使用了同一种判断溢出的方式，即在要添加最后一个数字之前判断其是否会超出当前
int所能表示的最大值。 - 代码本身还是比较直白的，直接看代码吧，这种题主要还是细节。
algorithm
/**
* Runtime: 4 ms, faster than 77.74% of C++ online submissions for Reverse Integer.
* Memory Usage: 8.1 MB, less than 77.50% of C++ online submissions for Reverse Integer.
*/
class Solution {
public:
int reverse(int x) {
/* convert to positive number */
unsigned tmp_x = x;
char flag = 0;
if (x < 0) {
flag = 1;
tmp_x = x == INT_MIN ? (unsigned)x : -x;
}
unsigned result = 0;
while (tmp_x) {
result = result * 10 + tmp_x % 10;
tmp_x /= 10;
/* check overflow */
if (tmp_x < 10 && tmp_x && result > UINT_MAX / 10) {
return 0;
}
}
if (flag) {
if (result > (unsigned)INT_MIN)
return 0;
return -result;
}
if (result > INT_MAX)
return 0;
return (int)result;
}
};
epilogue
连续两题整数题，解法类似，都还好，没那么多弯弯绕绕。