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.

 

解题思路：

注意题目中要求不能有额外空间

（回文数，算出x的倒置数s，比较s是否和x相等就行了。）

bool isPalindrome(int x) {
    if(x<0)
        return false;
    int num=x;
    int s=0,mod;
    while(num)
    {
        mod=num%10;
        s=s*10+mod;
        num=num/10;
    }
    return(s==x);
}

其他的代码的思路：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

class Solution {
public:
    bool isPalindrome(int x) {
        
        //negative number
        if(x < 0)
            return false;
            
        int len = 1;
        while(x / len >= 10)//失误点
            len *= 10;
            
        while(x > 0)    {
            
            //get the head and tail number
            int left = x / len;
            int right = x % 10;
            
            if(left != right)
                return false;
            else    {
                //remove the head and tail number
                x = (x % len) / 10;
                len /= 100;
            }
        }
        
        return true;
    }
};