[LeetCode]Palindrome Number

不需要用reverse或者转换成string的方法。每次取头尾两个数比较即可，常数空间，linear time，代码如下：

	class Solution {
	public:
	bool isPalindrome(int x) {
	if(x < 0)return false;
	int div = 1;
	while(x / div >= 10)
	div *= 10;
	while(x)
	{
	int hi = x / div, lo = x % 10;
	if(hi != lo)
	return false;
	x %= div;
	x /= 10;
	div /= 100;
	}
	return true;
	}
	};

view raw Palindrome Number.cpp hosted with ❤ by GitHub