# 思路

將 x 顛倒，再判斷是否與原數相同

# 參考程式碼

	static auto fast_io = []
	{
	ios::sync_with_stdio(false);
	cout.tie(nullptr);
	cin.tie(nullptr);
	return 0;
	}();

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

	long x1 = x, x2 = 0;

	for (; x; x /= 10) x2 = x2*10 + x%10;
	return x1 == x2;
	}
	};