C# 写 LeetCode easy #7 Reverse Integer
7、Reverse Integer
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.
代码:
static void Main(string[] args) { int num = 123; int reversenum = Reverseinteger(num); Console.WriteLine(reversenum); Console.ReadKey(); } private static int Reverseinteger(int num) { var res = 0L; while (num != 0) { res = res * 10 + num % 10; num = num / 10; } if (res > int.MaxValue || res < int.MinValue) { res = 0; } return (int)res; }
解析:
输入:整数
输出:整数
代码思想:
首先,对非0整数从个位数开始向左循环,每次要得到的数都是上一次结果乘10的基础上加这一位的数,初始时上一次结果为0。每进行一次,该整数都将除以10,意思是不再考虑个位上的数。
其次,若溢出则返回0。否则返回最终的结果。
时间复杂度:O(n)