检查数字是否是回文 [英] Checking if a number is a palindrome

问题描述

我尝试使用以下代码检查数字是否为回文式:

I've tried to check whether a number is a palindrome with the following code:

unsigned short digitsof (unsigned int x)
{
    unsigned short n = 0;
    while (x)
    {
        x /= 10;
        n++;
    }
    return n;
}

bool ispalindrome (unsigned int x)
{
    unsigned short digits = digitsof (x);

    for (unsigned short i = 1; i <= digits / 2; i++)
    {
        if (x % (unsigned int)pow (10, i) != x % (unsigned int)pow (10, digits - 1 + i))
        {
            return false;
        }
    }
    return true;
}

但是，以下代码无法检查回文-即使数字是回文，也总是返回false.

However, the following code isn't able to check for palindromes - false is always returned even if the number is a palindrome.

任何人都可以指出错误吗?

Can anyone point out the error?

(请注意:我不希望将其变成字符串并反转它以查看问题出在哪里:相反，我有兴趣知道上面代码中的错误在哪里.)

(Please note: I'm not interested to make it into a string and reverse it to see where the problem is: rather, I'm interested to know where the error is in the above code.)

推荐答案

问题是这样的:

x % (unsigned int)pow (10, i)

尝试:

x =504405
i =3

SO I want 4.

x % 10^3 => 504405 %1000 => 405 NOT 4

怎么样

x / (unsigned int)pow (10, i -1) % 10

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