LeetCode算法入門- Reverse Integer-day6

LeetCode算法入門- Reverse Integer-day6

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

class Solution {
    public int reverse(int x) {
        long result = 0;
        //取絕對值
        int temp = Math.abs(x);
        while(temp > 0){
            //核心是這裏，取反的關鍵
            result = result * 10 + temp % 10;
            temp = temp / 10;
        }
        //記得判斷是否溢出，這裏很關鍵，這道題的關鍵
        if(result > Integer.MAX_VALUE)
            return 0;
        if(x >= 0)
            return (int)result;
        //這裏的符號是另外來進行判斷，還有就是強制類型轉換，將long強轉爲int
        else return -(int)result;
    }
}

方法二：將int類型轉換爲StringBuffer類型，然後調用reverse()方法

class Solution {
    public int reverse(int x) {
        int temp = 0;
        //這裏是原因在下面圖片
        if(x == -2147483648)
            return 0;
            // temp = -x;
        else
            temp = Math.abs(x);

        String str = Integer.toString(temp);
        //記得要轉換爲StringBuffer類型纔行
        StringBuffer sb = new StringBuffer(str);
        String reverseStr = sb.reverse().toString();
        //將字符串類型轉換爲長整型（以防它出界）
        long l = Long.parseLong(reverseStr);
        if(l > Integer.MAX_VALUE)
            reverseStr = "0";
        if(x > 0)
            return Integer.parseInt(reverseStr);
        else
            return -Integer.parseInt(reverseStr);
    }
}