問題

palindrome-number

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Follow up: Could you solve it without converting the integer to a string?

Example 1:

Input: x = 121
Output: true

Example 2:

Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Example 4:

Input: x = -101
Output: false

Constraints:

• $-2^{31}<=x<=2^{31}-1$

思考

判斷是否是迴文數字,輸入是int.

首先,所有的負數都肯定不是. 然後,可以通過某種方式,依次取得最後和最前的數字進行比較.

比如, 1 2 3 4 2 1, 第一次比較1 和1 ,然後比較2 和2 然後比較3 和4 發現不相等,則返回false.

數字取最後一位很好做, 除10取餘數就可以了, 但是取最前一位需要知道最大位數.

想複雜了. 最簡單的辦法是把數字倒過來, 去判斷倒過來的這個數和輸入的數是否相等.

方法1

時間複雜度$O(1)$
空間複雜度$O(1)$.

class Solution {
    public boolean isPalindrome(int x) {
    	// 小於0 返回false
        if(x < 0)
            return false;
        
        // 計算 x 按位倒裝的數 r
        int r = 0,temp = x;
        while(temp != 0){
            r = r*10 + temp%10;
            temp /= 10;
        }
        return r == x;
    }
}