[Leetcode] Palindrome Number

最新推荐文章于 2020-08-18 15:23:01 发布

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本文介绍了一种不使用额外空间判断整数是否为回文数的方法，通过反转整数并比较原始值与反转后的值来实现。考虑到整数限制，此算法复杂度为O(1)。

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题目：

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.

思路：先将数字reverse，然后再比较是否相等。

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) return false;
        int original = x;
        long long reverse = 0;
        while (x > 0) {
            reverse = reverse * 10 + x % 10;
            x /= 10;
        }
        return original == reverse;
    }
};

总结：考虑到int输入只有32位，复杂度为O(1).