LeetCode9_Palindrome Number

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My answer

   /**     * case1: x is negative, x is definitely not palindrome<br>     * <del>case2: x is between 0 and 9, x is palindrome</del>.included by case3 <br>     * case3: x has more than 1 digit.     *     Algorithm:     *      step1:calculate the number of digits in X;     *      step2: if x.len is odd, compare left-right digit;      *             if x.len is even, compare mid-1 and mid digit before compare left-right digit,     */     public boolean isPalindrome(int x) {</span>        if (x < 0) {            return false;        }        int len = 1;        int quotient = x;        while (quotient / 10 > 0) {            len++;            quotient = quotient / 10;        }        int mid = len / 2;        if (len % 2 == 0) {            int leftDigit = x / (int) Math.pow(10, len - mid) % 10;            int rightDigit = x / (int) Math.pow(10, len - 1 - mid) % 10;            if (leftDigit != rightDigit) {                return false;            } else {                return isPalindrome(x, len, mid - 2, mid + 1);            }        } else {            return isPalindrome(x, len, mid - 1, mid + 1);        }    }    private boolean isPalindrome(int x, int len, int left, int right) {        while (left >= 0 && right < len) {            int leftDigit = x / (int) Math.pow(10, len - 1 - left) % 10;            int rightDigit = x / (int) Math.pow(10, len - 1 - right) % 10;            if (leftDigit != rightDigit) {                return false;            }            left--;            right++;        }        return true;    }

Summary

1. 负数例如-3,-123,一定不是palindrome

2. 不能把integer 转为string，因为题目要求without extra space

3. 不能将integer reverse后再判断，因为reverse可能会out of range,例如INT_MAX=2147483647, reverse后溢出了。

4. 如何计算integer x有多少位？

        int len = 1;        int quotient = x;        while (quotient / 10 > 0) {            len++;            quotient = quotient / 10;        }

5. 如何得到每一位数？

先把该位移动到个位，再%10，见方法：private boolean isPalindrome(int x, int len, int left, int right)

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