【LittleDuck_LeetcodeNote】06 - Palindrome Number

【Instagram】 / 【巴哈姆特小屋】 / 【Vocus方格子】 / 【Medium】
喜歡的話就按個愛心或贊助ㄅ。

Question and Hints:

Given an integer x, return true if x is a palindrome, and false otherwise.

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

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.

Constraints:

• 231 <= x <= 231 - 1

Follow up:

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

First Solution:

💡 先將num轉成文字後切割，再用for loop同時從最前面和最後面檢查，
只要有不同的就回傳false，
若for loop順利跑完，表示符合palindrome，return true.

class Solution {
    public boolean isPalindrome(int x) {
        String[] num = String.valueOf(x).split("");
        for(int i = 0; i < num.length; i++){
            int end = num.length-1 - i;
            if(end - i > 0){
                if(!(num[i].equals(num[end])))
                    return false;
            }
        }
        return true;
    }
}

⏰ Runtime: 29 ms (9.89 %)

📦 Memory: 42 MB (55.87 %)

Upgrade Solution:

think of a math problem.

💡 先設定一些一定不符合迴文的條件：

1. 負數
2. 大於0且尾數為0

以12321和123321為例，
理論上如果是迴文，從頭到中間值的組成，應該和從尾到中間值的組成一樣，
都是123，1x10x10 + 2x10 + 3，
畫一遍就會很清楚ㄌ，有點像用check從後面往前吃掉一半的感覺。

class Solution {
    public boolean isPalindrome(int x) {
        if(x<0 || x!=0 && x%10 ==0)
	　　return false;
        int check=0;
        while(x>check){
	    check = check*10 + x%10;
          　x/=10;
        }
        return (x==check || x==check/10);
    }
}

⏰ Runtime: 9 ms (99.32 %)

📦 Memory: 42 MB (65.91 %)