leetcode[9]Palindrome Number

本文介绍了一种不使用额外空间来判断整数是否为回文数的方法,通过比较整数从两边开始的数字来实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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.

class Solution {
public:
    bool isPalindrome(int x) {
    if (x<0)return false;
    if (x==0)return true;
    long long base=1;
    while(x/base)
    {
        base*=10;
    }
    base/=10;
    int left=0;
    int right=0;
    while(0!=x)
    {    
        left=x/base;
        right=x%10;
        if (left!=right)
        {
            return false;
        }
        else
        {
            x-=base*left;
            x/=10;
            base/=100;
        }
    }
    return true;
    }
};

 

转载于:https://www.cnblogs.com/Vae1990Silence/p/4283700.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值