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LeetCode 第九题, Palindrome Number

2019年03月02日 ⁄ 综合 ⁄ 共 1167字 ⁄ 字号 评论关闭

题目原文:

Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

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.

题意解析:

即是判断一个数字是否是回文串.

水题~~~~~~~~~~~~~~~~~~~~~~~~~

代码如下:

solution 1:依次取出左右的数字,对比.

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;
        if (x == 0)
            return true;
            
        int base = 1;
        while(x / base >= 10)
            base *= 10;
            
        while(x)
        {
            int leftDigit = x / base;
            int rightDigit = x % 10;
            if (leftDigit != rightDigit)
                return false;
            
            x -= base * leftDigit;
            base /= 100;
            x /= 10;
        }
        
        return true;
    }
};

solution 2: 将数字取反....判断是否相等,不过,翻转的时候可能会溢出,但是,也是过了..............

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;
        
        long long target = x;
        //if(target < 0) target = -target; 
        long long reverse = 0;
        long long curNum = target;
        while(curNum != 0)
        {
            reverse = 10*reverse+curNum%10;
            curNum /= 10;
        }
        
        return reverse == target;
    }
};

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