[LeetCode]Palindrome Number

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。通过算法直接操作整数,避免了转换为字符串的过程,同时考虑了负数和溢出等特殊情况。

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

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.


/*
	题号:
    Determine whether an integer is a palindrome. Do this without extra space.
*/
#include<iostream>
using namespace std;


class Solution{
public:
	bool ispalindrome(int x){
		if (x < 0)
			return false;
		if (x == 0)
			return true;

		int base = 1;
		while (x / base >= 10)
			base *= 10;  //base为10^n,表示取出数的最高位

		while (x){
			int leftDigit = x / base;
			int rightDigit = x % 10;
			if (leftDigit != rightDigit)
				return false;

			x -= base * leftDigit;
			x /= 10;
			
			base /= 100;
		}
		return true;

	}
};

int main()
{
	Solution solution;
	int n1 = -123321;
	int n2 = 12321;
	int n3 = 12341;
	int n4 = 1001;

	cout << solution.ispalindrome(n1) << endl;
	cout << solution.ispalindrome(n2) << endl;
	cout << solution.ispalindrome(n3) << endl;
	cout << solution.ispalindrome(n4) << endl;

	getchar();
	return 0;
}


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