LeetCode 7. Reverse Integer(数学)
LeetCode 7 Reverse Integer数学
问题描述解题思路参考代码
By ScarbScarb’s Blog
Tags: - Math
问题描述
Reverse digits of an integer.
Example1: x = 123, return 321 Example2: x = -123, return -321
Note: The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
解题思路
题意是将一个整数转置输出。提示：转置输出的数溢出时返回0. 所以要处理一下转置后的数溢出的特殊情况，这里包括了正溢出和负溢出。
参考代码
#include <iostream>
using namespace std;
class Solution {
public:
int reverse(
int x) {
int num =
0;
while (x)
{
if (
abs(num) > INT_MAX /
10 ||
(
abs(num) == INT_MAX /
10 &&
abs(x) %
10 > INT_MAX %
10))
return 0;
num = num *
10 + x %
10;
x /=
10;
}
return num;
}
};
int main()
{
int i =
123;
Solution sl = Solution();
cout << sl.reverse(
123) << endl;
cout << sl.reverse(-
123) << endl;
cout << sl.reverse(
103) << endl;
cout << sl.reverse(
120) << endl;
cout << sl.reverse(
1000000003) << endl;
cout << sl.reverse(-
1000000003) << endl;
system(
"pause");
return 0;
}
