【leetcode刷題筆記】Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

解題：設定一個變量sum存放反轉後的答案，每次取輸入x的最後一位n，並用sum = sum*10+n更新sum。

例如題設的數字123，初始sum爲0，過程以下：

取末尾的3，sum=3，x=12

取末尾的2，sum=32，x=1

取末尾的1，sum=321，x=0

特別注意x一開始就是0的處理，直接返回0就能夠了。

代碼：

 1 class Solution {
 2 public:
 3     int reverse(int x) {
 4         int sum = 0;
 5         if(x == 0)
 6             return 0;
 7         while(x != 0){
 8             int temp = x%10;
 9             sum = sum*10+temp;
10             x /= 10;
11         }
12         return sum;
13     }
14 };

題目還給出了一些思考：

1.If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

  上述的方法避免了這個問題，10和100的輸出都是1

2.Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases? Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).