[leetcode] 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.

https://oj.leetcode.com/problems/palindrome-number/

思 路1（不推荐）：利用Reverse Integer的方法，求的转换后的数字，然后比较是否相等。提示说这样有溢出的问题，想了想感觉问题不大，leetcode也过了。因为首先输入必须是 一个合法的int值，负数直接返回false，对于正数，假设输入的int是一个palindrome，reverse之后依然不会溢出，所以正常返回 true；所以如果转换后溢出了，证明肯定不是palindrome，溢出后的数字跟输入一般不相同（想不出相等的情况-_-!，求指点），所以返回了 false。

思路2：从两头依次取数字比较，向中间推进。

public class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0)
            return false;
        //calcu the length of digit
        int len = 1;
        while (x / len >= 10) {
            len *= 10;
        }

        while (x != 0) {
            int left = x / len;
            int right = x % 10;

            if (left != right)
                return false;
            //remove the head and tail digit
            x = (x % len) / 10;
            len /= 100;
        }

        return true;
    }

}

 第二遍记录：

　　注意如何取第一个数字和最后一个数字

　　注意divis的计算方法和位数。

public class Solution {
    public boolean isPalindrome(int x) {
        if(x<0)
            return false;
        int divis=1;
        while(x/divis>=10){
            divis *=10;
        }
    
        while(x!=0){
            int first = x/divis;
            int last = x%10;
            if(first!=last)
                return false;
            
            x = (x%divis)/10;
            divis /=100;
        }
        return true;
    }
}

参考：

http://www.programcreek.com/2013/02/leetcode-palindrome-number-java/

http://fisherlei.blogspot.com/2012/12/leetcode-palindrome-number.html