#leetcode刷题之路29- 两数相除

给定两个整数，被除数 dividend 和除数 divisor。将两数相除，要求不使用乘法、除法和 mod 运算符。
返回被除数 dividend 除以除数 divisor 得到的商。

示例 1:
输入: dividend = 10, divisor = 3
输出: 3
示例 2:
输入: dividend = 7, divisor = -3
输出: -2

int divide(int dividend, int divisor) {

    int count=;
    if((dividend>INT32_MAX||dividend<INT32_MIN)||(divisor>INT32_MAX||divisor<INT32_MIN)) return INT32_MAX;
    if((dividend)==INT32_MIN&&(divisor==INT32_MIN)) return  ;
    if((dividend)==INT32_MIN&&(divisor==)) return INT32_MIN;
    if(divisor==INT32_MIN) return ;
    if((dividend)==INT32_MIN&&(divisor==-)) return INT32_MAX;
    if(dividend==INT32_MIN)
    {
        bool flag=;
        if(divisor<) flag=;
        divisor=(divisor>)?divisor:-divisor;
        while(dividend<=)
        {
            dividend+=divisor;
            count++;
        }
        return flag?count-:-count;
    }
    if(dividend==||abs(dividend)<abs(divisor)) return count;
    if(divisor==) return dividend;
    if(divisor==-) return -dividend;
    else if(dividend>&&divisor>||dividend<&&divisor<)
    {
         dividend=abs(dividend);
        divisor=abs(divisor);
        while (dividend>=)
        {
            dividend-=divisor;
            count++;
        }
        return (count-);
    }
    else if(dividend<&&divisor>||dividend>&&divisor<)
    {
        dividend=(dividend>)?dividend:-dividend;
        divisor=(divisor>)?divisor:-divisor;
        while (dividend>=)
        {
            dividend-=divisor;
            count++;
        }
        return  -count;
    }
    return ;
}