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Palindrome Number

#algorithms #csharp #leetcode #easy

#9.Palindrome Number

Problem statement

Given an integer x, return true if x is palindrome integer.

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

  • For example, 121 is a palindrome while 123 is not.

Example 1

Input: x = 121 Output: true Explanation: 121 reads as 121 from left to right and from right to left.
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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.
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Example 3

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

給定一個 x 的整數，如果是回文則返回 true，否則 false

回文的定義是不論從頭或從尾開始，讀起來是一樣的，例如 121 和 1221 是回文，但 123 不是

Solution

這題要考慮的地方是回文的長度，偶數長度的回文分割為左右兩等分時比較好判斷，奇數長度的回文最後會留下中間的數，它不需與其他數對應，且題目最後的 Follow up 還提到能否在不將數字轉換成字串的情況下來解，依照前述的分析及題意，此題可使用算術運算子（Arithmetic operators）來解。

一開始先過濾一些特定的條件

  • 判斷 x 是否為 0，如果是 return true
  • 判斷 x 是否小於 0 或者為 10 的倍數，如果是 return false

接著執行 while 迴圈直到條件成立，最後 return x 與擷取出的 reverseNum 是否相等。

public bool IsPalindrome(int x) { if (x == 0) return true; if (x < 0 || x % 10 == 0) return false; int reverseNum = 0; while (x > reverseNum) { reverseNum = reverseNum * 10 + x % 10; x /= 10; } return x == reverseNum || x == reverseNum / 10; }
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Reference

LeetCode Solution

GitHub Repository

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