LeetCode 009 Palindrome Number

class Solution {
public:
	int _size(int x){
		int size = 0;
		int tx = x;
		while(tx){
			tx /= 10;
			size++;
		}
		return size;
	}
    bool isPalindrome(int x) {
		if(x<0)return false;
		else if(x>=0 && x<=9)return true;
		else{
			int size = _size(x);
			int head = x/(int)pow(10,size-1);
			int tail = x%10;
			if(head != tail){
				return false;
			}
			x = x%(int)pow(10,size-1);
			x = x/10;
			int newsize = _size(x);
			if(size-newsize > 2)
				if(x%(int)pow(10, size-newsize-2) == 0) //考虑数后跟着一串0的情况
					x = x/(int)pow(10, size-newsize-2);
				else
					return false;
			return isPalindrome(x);
		}
    }
};

【思路2】

【代码】

class Solution {
public:
    bool isPalindrome(int x) {
	    long long reverseX=0;
	    if(x<0)return false;
	    int xcopy=x;
	    while(xcopy){       //将整数倒置
	        reverseX=reverseX*10+xcopy%10;
	        xcopy/=10;
	    }
	    if(reverseX>INT_MAX)return false;   //判断倒置后的结果是否越界
	    if(x==(int)reverseX)return true;      //判断倒置结果和原数是否相等
	    return false;
    }
};