c++ - Checking Palindrome numbers in C++

Question

I'm trying to use C++ to solve Project Euler problem 4.

I thought of placing each digit into an array and seeing if the values of the elements are the same in the correct places. This is what I have so far:

int digits[]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

But I can't write the code. How can I take the number and make it seem that way? I know the basics of C++ (loops and pointers). How to use adjustable array in C++ and should I use this? I also don't know what the size of the array should be. Is there another better algorithm to implement?

Answer 1

If you know array/string length you can loop and compare element [i] with element [length-i-1]

Answer 2

您可以使用循环来检查两端是否匹配：

int digits[]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
int length = sizeof(digits)/sizeof(digits[0]);
for (int i = 0; i < length; i++) {
    if (digit[i] != digits[length - i - 1])
        return false; // at this point, you know the number is not a palindrome 
return true;

Answer 3

长度计算有点类似于 C 的方法，但它确实有效。

#include <iostream>
using namespace std;

bool isPalindrome(int* digit_array, int len) {
    if(len <= 1) return true;
    if(digit_array[0] != digit_array[len-1])
        return false;
    else
        return isPalindrome(++digit_array, len-2);
}

int main() {
    int digits[] = {1, 2, 3, 4};
    int digits2[] = {1, 2, 2, 1};

    // it returns false
    cout << "isPalindrome(digits) => " << isPalindrome(digits, (sizeof(digits)/sizeof(*digits))) << endl;

    // it returns true
    cout << "isPalindrome(digits2) => " << isPalindrome(digits2, (sizeof(digits2)/sizeof(*digits2))) << endl;

    return 0;
}

编辑：我刚刚阅读了 Project Euler 问题 4。也许递归方法不是实现它的最佳方法。

Answer 4

我会将整数转换为字符串并检查字符串的反转是否等于自身。

#include <iostream>
#include <string>

int main()
{
    int num = 21334;
    auto s = std::to_string( num );

    if (std::string(s.begin(), s.end()) == std::string(s.rbegin(), s.rend()))
    {
        std::cout << "num is a palindrome";
    } else
    {
        std::cout << "num is not a palindrome";
    }
}

Answer 5

我希望你现在已经解决了这个问题，但是对于那些最近被难住的人来说，有一种更简单的方法可以在不弄乱字符串的情况下完成这个问题。您可以结合使用 mod 函数和除法（% 和 /）来单独查看每个数字。我解决这个问题的方法是向后构建数字，然后比较它们。这里的关键是要知道 number%10 会给你个位数字，然后 number/10 会砍掉个位数字。玩得开心，我真的很喜欢解决一些体育问题！