[LeetCode]- Reverse Integer（將一個整數反轉）

[ 問題: ]

Reverse digits of an integer. Example1: x = 123, return 321
Example2: x = -123, return -321

 

題解：給定一個整數，將這個整數的數字旋轉位置。

[ 陷阱 : ]

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
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). 注意點1：如果一個整數最後一位位0，比如：10100，那麼反轉後將變成101。 注意點2：如果輸入的數為1000000003，那麼反轉後將會出現溢出，需要注意。
[ 解法 : ]

/*
 * 假設輸入123:
 * 123 -> 3  
 * 12 -> 30 + 2  
 * 1 -> 320 + 1  
 */
public class Solution {
	public int reverse(int x) {
		int result = 0;
		while (x != 0) {
			result = result * 10 + x % 10; // 每一次都在原來結果的基礎上變大10倍，再加上余數
			x = x / 10; // 對x不停除10
		}
		return result;
	}

	public static void main(String[] args) {
		Scanner scanner = new Scanner(new BufferedInputStream(System.in));
		while (scanner.hasNext()) {
			int num = scanner.nextInt();
			System.out.println(new Solution().reverse(num));
		}
	}
}