换了n种方法，终于是跑通了一个不超时的，记录一下

class Solution {
    public int divide(int dividend, int divisor) {
        boolean isMinus = ((dividend<0 && divisor >0) || (dividend>0 && divisor <0)); //异号为true，同号为false
        if( Integer.MIN_VALUE == dividend){
            if(1 == divisor)
                return dividend;
            else if(2 == divisor)
                return dividend >> 1;
            else if(Integer.MIN_VALUE == divisor)
                return 1;
        }else if(Integer.MIN_VALUE == divisor)
            return 0;
        dividend = Math.abs(dividend);
        //abs(Integer.MIN_VALUE) = Integer.MIN_VALUE
        dividend = dividend == Integer.MIN_VALUE ? Integer.MAX_VALUE : dividend;
        divisor = Math.abs(divisor);
        int sub = divisor; //接下来要减的数，每次翻倍
        int c = 1;//减的数翻倍，要加到结果上的数也得翻倍
        int ret = 0;
        while(dividend >= divisor){
            if(dividend >= sub){
                dividend=minus(dividend, sub);
                ret=add(ret,c);
                sub=(sub<<1);
                c=(c<<1);
            }else{
                sub=(sub>>1);
                c=(c>>1);
            }
        }
        if(isMinus)
            ret=add(~ret,1);
        return ret;
    }
    public int multi(int a, int b){
        if(b == 0)
            return 0;
        else if(1 == b)
            return a;
        else if(-1 == b)
            return 0-a;
        if(a < 0 && b < 0){
            return multi(add(~a, 1), add(~b, 1));
        }else if(a < 0 && b > 0){
            return 0 - multi(add(~a, 1), b);
        }else if(a > 0 && b < 0){
            return 0 - multi(a, add(~b, 1));
        }else{
            for(int i = 1; i < 32; i++){
                if( 1 << i > b){ //找到比b大的且是2的幂次的最小数
                    int j = 1 << (i-1); //比b小的且是2的幂次的最大数
                    return add(a << (i-1) , multi(a, b-j));
                }
            }
        }
        return 0;
    }
    public int minus(int a, int b){
        return add(a, add(~b, 1));
    }
    public int add(int a, int b){
        int aa = a;
        int bb = b; //进位
        while(bb != 0){
            aa = a ^ b;
            bb = (a & b) << 1;
            b = bb;
            a = aa;
        }
        return aa;
    }
}