金城式的程式筆記

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.

題目:

　　給一個32位元的整數，把它反轉過來。

注意:

　　integer的範圍為 −2³¹,  2³¹ − 1 ，如果反轉溢出的話則回傳0。

解答:

public class ReverseInteger7 {
    public static void main(String[] args) {
        System.out.print(reverse(123));
    }
    public static int reverse(int x) {

//考慮反轉後可能溢位故用長整數來儲存
        long r = 0;
        while (x != 0) {

//把x的最小位數取出,反轉成r的最大位數
            r = r * 10 + x % 10;
            x = x / 10;
        }

//判斷是否會溢位
        if (r > Integer.MAX_VALUE || r < Integer.MIN_VALUE) {
            return 0;
        } else {
            return (int)r;
        }
    }
}