LeetCode(35):Palindrome Number 分类: leetCode 2015-07-10 09:26 161人阅读 评论(0) 收藏

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.

       判断一个数是不是回文数,首先负数不可能是回文数;其次要求不能用“额外”的存储空间,这句话的意思并非表示不能用存储空间,应该是表示不能用过多的存储空间;那么可以判断首位与尾位是数值是否相等,再去掉头尾,再次判断首位和尾位是否相等,直至该值为0为止。


class Solution {
public:
    bool isPalindrome(int x) {
       	int a = static_cast<int>(floor(log10(x)) + 1);	
    	int b= static_cast<int>(pow(10, a - 1));
    
    	if (x < 0)
    		return false;
    	while (x)
    	{
    		if (x % 10 != x / b)return false;
    				
    		x = x % b;//去掉最高位
    		x = x / 10;//去掉最低位
    		//注意,以上两够不可逆,取余不一定能去掉最后一位
    		b = b / 100;
    	}
    	return true;
    }
};



原文地址:https://www.cnblogs.com/zclzqbx/p/4687064.html