Leetcode 9. Palindrome Number(判断回文数字）

Determine whether an integer is a palindrome. Do this without extra space.（不要使用额外的空间）

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.

分析：

回文数，并且题目要求不能使用额外的空间。

即，不能使用回文串的方法。

本题很简单，算出x的倒置数a，比较a是否和x相等就行了。

 1 class Solution {
 2 public:
 3     bool isPalindrome(int x) {
 4         //算出x的倒置数a，比较a是否和x相等就行了
 5         int a = 0, b = x;
 6         while(b > 0){
 7             a = a * 10 + b % 10;
 8             b /= 10;
 9         }
10         if(a == x)
11             return true;
12         else
13             return false;
14
15     }
16 };

网上其他的代码，其思路：每次提取头尾两个数，判断它们是否相等，判断后去掉头尾两个数

 1 class Solution {
 2 public:
 3     bool isPalindrome(int x) {
 4
 5         //negative number
 6         if(x < 0)
 7             return false;
 8
 9         int len = 1;
10         while(x / len >= 10)
11             len *= 10;
12
13         while(x > 0)    {
14
15             //get the head and tail number
16             int left = x / len;
17             int right = x % 10;
18
19             if(left != right)
20                 return false;
21             else    {
22                 //remove the head and tail number
23                 x = (x % len) / 10;
24                 len /= 100;
25             }
26         }
27
28         return true;
29     }
30 };