[LeetCode] Palindrome Number

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

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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.

两种方法：

第一种：借助之前实现的reverse integer的方法，将int x reverse 得到 y，然后比较x，y是否相等，

第二种：先得到x的base，注意，1231的base是10^3,然后1234/10^3得到1，和最后的1作比较，然后x=1234-1*1000 = 234，x=x/10=23，base=base/100=10,

　　　　反复迭代，可以得到答案

 参考http://www.cnblogs.com/remlostime/archive/2012/11/13/2767676.html

class Solution {
    public:
        int reverse(int x) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            int sign = x > 0 ? 1 : -1;
            long long y = abs(x);

            long long ret = 0;
            while(y)
            {   
                int digit = y % 10; 
                ret = ret * 10 + digit;
                y /= 10; 
                if(ret * sign > INT_MAX)
                    return 0;
                if(ret * sign < INT_MIN)
                    return 0;
            }   

            return ret * sign;
        }   
#if 0
        bool isPalindrome(int x) {
            if(x == 0)
                return true;
            if(x <0)
                return false;
            int y = reverse(x);
            if(x == y)
                return true;
            else 
                return false;
        }  
#endif        
    bool isPalindrome(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (x < 0)
            return false;
        if (x < 10)
            return true;
            
        int base = 1;
        while(x / base >= 10)
            base *= 10;
            
        while(x)
        {
            int leftDigit = x / base;
            int rightDigit = x % 10;
            if (leftDigit != rightDigit)
                return false;
            
            x -= base * leftDigit;
            base /= 100;
            x /= 10;
        }
        
        return true;
    }
};