9. Palindrome Number

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.

Normal Way

reverse the Integer with long Integer

1 2 3 4 5 6 7 8 9 10 11 12 13 14

class  Solution { public :      bool  isPalindrome( int  x) {          if (x < 0)  return  false ;          long  num = x,result = 0;          while (num){              result = result * 10 + num %10;              num/=10;          }          if ( result == x)  return  true ;          else           return  false ;      } };

Improved Way

What if the input number is already a long integer?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

class  Solution { public :      bool  isPalindrome( int  x) {          if (x<0)  return  false ;          if (x<10)  return  true ;          if (x%10==0)  return  false ;                   //actual logic          int  v=x%10;          x=x/10;          while (x-v>0) //x is getting smaller every step,          {                  v=v*10+x%10;                  x/=10;          }          if (v>x){v/=10;}          return  v==x? true : false ;      } };

比如 

奇数长度：1234321

每次通过除以10和对10取余来保存下前半部分x和后半部分v，比如 x = 123432, v=1;

直到x=123，v=1234的时候，循环结束，再通过 v/10 == x，来判断数是不是回环的。

偶数长度：则直接比较 v==x 

Genetic Way

compare the first and last number of the input x. 

Maximum comparing time will be half of the length of x

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

class  Solution { public :      bool  isPalindrome( int  x) {          if  (x < 0)  return  false ;          int  d = 1;  // divisor          while  (x / d >= 10) d *= 10;  // 计算出是多少位数          while  (x > 0) {              int  q = x / d;  // quotient              int  r = x % 10;  // remainder              if  (q != r)  return  false ;              x = x % d / 10;              d /= 100;          }          return  true ;      } };

来自为知笔记(Wiz)

转载于:https://www.cnblogs.com/zhxshseu/p/27ea3ddf3ceda4bbd5c090b514b28c13.html