Leetcode 9. Palindrome Number(判斷回文數字）

Determine whether an integer is a palindrome. Do this without extra space.（不要使用額外的空間）

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：

回文數，並且題目要求不能使用額外的空間。

 

即，不能使用回文串的方法。

 

本題很簡單，算出x的倒置數a，比較a是否和x相等就行了。

 

class Solution {
public:
    bool isPalindrome(int x) {
        //算出x的倒置數a，比較a是否和x相等就行了
        int a = 0, b = x;
        while(b > 0){
            a = a * 10 + b % 10;
            b /= 10;
        }
        if(a == x)
            return true;
        else
            return false;
        
    }
};

 

網上其他的代碼，其思路：每次提取頭尾兩個數，判斷它們是否相等，判斷后去掉頭尾兩個數

 

class Solution {
public:
    bool isPalindrome(int x) {
        
        //negative number
        if(x < 0)
            return false;
            
        int len = 1;
        while(x / len >= 10)
            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;
    }
};