lintcode 容易题：reverse integer 颠倒整数

public class Solution {
    /**
     * @param n the integer to be reversed
     * @return the reversed integer
     */
    public int reverseInteger(int n) {
        // Write your code here
        int MAX = Integer.MAX_VALUE;
        if(n>=0){
            int res = 0;
            int num = n;
            while(n!=0){
                if(res>MAX/10) return 0;
                res =res *10 + n%10;
                n = n/10;
            }
            return res;
        }else{
            int res = reverseInteger(-n);
            return -res;
        }
        
    }
}

class Solution:
    # @param {int} n the integer to be reversed
    # @return {int} the reversed integer
    def reverseInteger(self, n):
        # Write your code here
        flag = False
        if n<0:
            n = -n
            flag = True
        res = 0
        while n!=0:
            res = res * 10 + n%10;
            n = n/10
        if flag:
            return -res
        else:
            return res