Palindrome Number -- leetcode_palindrome number leetcode-优快云博客

本文探讨了如何不使用额外空间来判断一个整数是否为回文整数，包括新颖的反转数字方法及对溢出的处理策略。

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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.

题目检测一个整数是不是回文整数，如12321就是一个回文整数。

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) return false;

        int reverse = 0;
        int power = 1;
        while (x/power/10) {
                power *= 10;
                reverse = reverse * 10 + x/power%10;
        }
        return x%power == reverse;
    }
};

我的思路，是利用reverse number，外加对overflow的适当处理。

感觉还是比较新颖和独到。

以下收录了网上博客的常见写法，供大家比较：

方法一：

class Solution {
public:
    bool isPalindrome(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(x < 0) return false;
        
        int div = 1;
        while(x/10 >= div){     // get large division
            div *= 10;
        }
        
        while(x > 9){
            int high = x / div;     // left digit
            int low = x % 10;       // right digit
            
            if(high != low){
                return false;
            }
            
            x = (x % div) / 10;     // get number between first and last
            div /= 100;
        }
        
        return true;
    }
};

方法二（来源）：

class Solution {
public:
    bool check(int x, int &y) {
        if (x == 0) return true;
    
        if (check(x/10, y) && (x%10 == y%10)) {
            y /= 10;
            return true;
        } else {
            return false;
        }
    }

    bool isPalindrome(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(x < 0) return false;
        
        return check(x, x);
    }
};

方法二是比较有特色的。但严格意义上讲，它不满足题目要求。题目要求是不使用辅助空间。