LeetCode Palindrome Number

最新推荐文章于 2020-08-18 15:23:01 发布

原创最新推荐文章于 2020-08-18 15:23:01 发布 · 362 阅读

0 ·

CC 4.0 BY-SA版权

ACM 专栏收录该内容

13 篇文章

订阅专栏

本文探讨如何在不使用额外空间的情况下，判断一个整数是否为回文数。通过利用log10求出数字长度，并从高位和低位开始依次比较，实现回文数的判断。特别关注负数和0的特殊情况。

题目来源： https://leetcode.com/problems/palindrome-number/

Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

Some hints:

Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.

题目要求不能使用额外空间，那就只能使用log10求出数字长度，然后分别从高位和低位开始，依次判断是否想等了。

题目需要注意以下几点：

1. 负数直接返回false;

2. 0应该返回true;

代码如下：

class Solution {
public:
	bool isPalindrome(int x) {
		if (x < 0)
			return false;
		int len = log10(x);
		for (int i = 0; i < (len / 2 + 1); i++)
		{
			if ((x / (int)pow(10, i)) % 10 != (x / (int)pow(10, len - i)) % 10)
				return false;
		}
		return true;
	}
};