LeetCode:Palindrome Number

基本思路：把 int 转换成string  借用stringstream  时间复杂度O(n)

  

class Solution {
public:
   public:
    bool isPalindrome(int x) {

       stringstream ss;
       ss<<x;
       string str=ss.str();
       int sfront = 0;
       int sback=str.length()-1;
        while(sfront<sback)
        {
         if(str[sfront]!=str[sback])
            return false;
         ++sfront;
         --sback;
        }
        return true;
    }
};