回文数.cpp

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
	bool isPalindrome(int x)
	{
		string s = to_string(x);
		int left = 0, right = s.size() - 1;
		while (left < right)
		{
			if (s[left] == s[right])
			{
				++left;
				--right;
			}
			else
				return false;
		}
		return true;
	}

	// 这是进阶版，不将整数转化为string
	// 将整数翻转一半来进行比较
	bool isPalindrome_1(int x)
	{
		// 负数或者最后一位为0，都是false
		if (x < 0 || (x != 0 && x % 10 == 0))
			return false;
		int revertedNumber = 0;
		// 当反转后的数字大于x时，说明已经反转了一半了
		while (revertedNumber < x)
		{
			revertedNumber = revertedNumber * 10 + x % 10;
			x /= 10;
		}
		return x == revertedNumber || x == revertedNumber / 10;
	}
	
	//update 2019.3.1
	bool isPalindrome2(int x) 
	{
		if (x<0||(x!=0&&x%10==0))
		    return false;
		int num=0;
		while (x>num)
		{
		    int digit = x%10;
		    num = num*10+digit;
		    x /= 10;
		}
		return (num==x || num /10 == x);
    	}
	
	//update 2019.9.23
	bool isPalindrome3(int x) 
	{
		if (x < 0 || (x != 0 && x % 10 == 0))
			return false;
		if (x >= 0 && x < 10)
			return true;
		int res = 0;
		while (x>res)
		{
			int tmp = x % 10;
			res = res * 10 + tmp;
			x = x / 10;
			if (res == x || res == x / 10)
				return true;
		}
		return false;
	}
};


int main()
{
	Solution a;
	cout << a.isPalindrome_1(121) << endl;



	system("pause");

}