LeetCode 9. Palindrome Number （迴文字數字）

 

題目地址：https://leetcode.com/problems/palindrome-number/description/

 

題目要求：

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

Example 1:

Input: 121
Output: true

Example 2:

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

Follow up:

Coud you solve it without converting the integer to a string?

算法思路：

第一種思路：把數字轉化爲字符串，再經過字符來作。

 

• 負數不多是迴文字數字，直接返回false
• 經過left和right兩個指針分別從中間往兩邊走依次比較，若是兩個字符不一樣返回false
• left容易肯定，直接經過除2而後1便可（角標從0開始），若是是偶數right爲left+1，不然則right爲left+2

 

題目代碼：

class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        String strX = String.valueOf(x);
        int length = strX.length();
        int left = length / 2 - 1;
        int right = length % 2 == 0 ? left+1 : left + 2;
        while (left >= 0){
            if(strX.charAt(left) != strX.charAt(right)){
                return false;
            }
            left--;
            right++;
        }

        return true;
    }
}

11508 / 11508 test cases passed.
Status: Accepted
Runtime: 322 ms

第二種思路：直接經過數字的反轉來作

 

• 利用一個變量暫存初始的x
• 負數直接返回false
• 反轉字符串存入result，在此過程當中防止超過整數最大值
• 最後判斷反轉後的整數是否和原始整數相等

  public boolean isPalindrome(int x) {
        int y = x;
        if (x < 0) {
            return false;
        }

        int result = 0;
        while(x !=0){
            
             if (result*10 + x%10>Integer.MAX_VALUE){       
                 return false;
             }
            result = result*10 + x%10;
            x = x/10;
        }
        return result == y;
    }

 

11508 / 11508 test cases passed.	Status:  Accepted
Runtime:  268 ms