leetcode 9. Palindrome Number

    public boolean isPalindrome(int x) {
        //reverse integer,and check whether the same
        if(x<0) return false;
        int newx=0;
        int m=x;
        while(x!=0){
            newx=newx*10+x%10;
            x=x/10;
        }
        return m==newx;
        
       
    }

上述算法的缺点：循环次数多，优化：n ==》n/2

  public boolean isPalindrome(int x) {
        //计算reverse integer
        if(x<0||(x>0 && x%10==0)) return false;
        int newx=0;
        while(x>newx){
            newx=newx*10+x%10;
            x=x/10;
        }
        return (x==newx||x==(newx/10));
    }