[leet code] Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321

Example2: x = -123, return -321

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Have you thought about this?

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).

public class Solution {
    public int reverse(int x) {
        boolean neg = false;
        if (x < 0) {
            neg = true;
            x = -x;
        }
        int y = 0;
        while (x>0){
            y = y*10 + x%10;
            x = x/10;
        }
        return neg?-y:y;
    }
}
           

思路總結:

1. 利用求餘%逐次獲得int變量中的每一位

2. 判斷是否為負數

3. int長度為32位 從-2的31次方到2的31次方