Sep-28-2018

争取每周做五个LeedCode题，定期更新，难度由简到难

Title: Palindrome Number

Description:

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example:

Input: 121
Output: true

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Difficulty:

Easy

Note:

Could you solve it without converting the integer to a string?

Implement Programming Language:

C#

Answer:

我是通过求了一下长度，再通过运算对比两侧数字解决的。
看了下LeetCode的Solution，是通过把后半段翻转，然后对比两个结果是否相同，例如1221，把21换成12，然后12==12，认为相同，确实比我的解法要简单，这次就贴两个答案。
My Answer:

public static bool IsPalindrome(long x)
        {
            if (x < 0 ||( x%10==0 && x!=0)) return false;
            int length = 1;
            long y = x;
            while(y/10 != 0)
            {
                y /= 10;
                length++;
            }
            Console.WriteLine($"length is {length}");
            for (int i = 1; i <= length/2; i++)
            {
                int j = length - i;
                var left = x / (long)Math.Pow(10, i - 1) % 10;
                var right = x / (long)Math.Pow(10, j) % 10;
                Console.WriteLine($"left is {left},right is {right}");
                if (left != right)
                    return false;
            }
            return true;
        }

LeetCode Solution:

public static bool IsPalindrome(long x)
        {
            if (x < 0 || (x % 10 == 0 && x != 0))
            {
                return false;
            }

            long revertedNumber = 0;
            while (x > revertedNumber)
            {
                revertedNumber = revertedNumber * 10 + x % 10;
                x /= 10;
            }
            return x == revertedNumber || x == revertedNumber / 10;//后半段解决12321这种奇数个数字的问题。
        }

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