LeetCode #Palindrome Number#

又是个软柿子啊．．．（主要是今天不知道在哪儿看到一个回文的题目了，然后就特地去LeetCode找了一下，还真有，一次性AC的感觉简直不能再爽）

我的Python版本解答:

Programmer  :   EOF
E-mail      :   jasonleaster@gmail.com
Date        :   2015.04.06
File        :   pn.py


class Solution:
    def isPalindrome(self, x):
        string = str(x)
        length = len(string)
        for i in range(0, length):
            if string[i] != string[length - i - 1] :
                return False
        
        return True

#----------- just for testing ----------


s = Solution()
if s.isPalindrome(123321) :
    print is palindrome

下面是皓神的C++解答:

皓神用了两种方法做解答．．．写了两个isPalindrome()的实现．

// Source : https://oj.leetcode.com/problems/palindrome-number/
// Author : Hao Chen
// Date   : 2014-06-18

/********************************************************************************** 
* 
* 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.
* 
*               
**********************************************************************************/

#include 
  
   

class Solution {
public:

    bool isPalindrome(int x) {
        if (x<0) {
            return false;
        }
        
        int len=1;
        for (len=1; (x/len) >= 10; len*=10 );
        
        while (x != 0 ) {
            int left = x / len;
            int right = x % 10;
            
            if(left!=right){
                return false;
            }
            
            x = (x%len) / 10;
            len /= 100;
        }
        return true;
    }
    
    bool isPalindrome2(int x) {
        return (x>=0 && x == reverse(x));
    }

private:    
    int reverse(int x) {
        int y=0;

        int n;
        while( x!=0 ){
            n = x%10;
            y = y*10 + n;
            x /= 10;
        }
        return y;
    }
};



int main()
{
    Solution s;
    printf(%d is %d
, 0, s.isPalindrome(0) );
    printf(%d is %d
, -101, s.isPalindrome(-101) );
    printf(%d is %d
, 1001, s.isPalindrome(1001) );
    printf(%d is %d
, 1234321, s.isPalindrome(1234321) );
    printf(%d is %d
, 2147447412, s.isPalindrome(2147447412) );
    printf(%d is %d
, 2142, s.isPalindrome(2142) );
}