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));    }