leet code -- Divide Two Integers

Q: Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT.

分析:設被除數爲p,除數爲q.則有:
p = q * (1+2+4+...+ 2^(exp2-1)) + new_p
每一輪循環除數翻倍增加:q=q+q,p=p-q直到最後的p小於除數q,此時被除數還剩new_p.
將(new_p,q)進行遞歸操作.
遞歸的出口是:p=0.
注意:需要注意的是,int類型範圍是[-2^31, 2^31-1],最高位爲補碼形式符號位!
當被除數p=-2^(31)=0x1<<31時,divide(-2^31, -1) = 2^31 - 1!! 0x1<<31默認int類型,值爲-2^31.

class Solution {
public:
    int divide(long long dividend, long long divisor) {                                                                               
        int exp2 = 0;
        long long ret;
        long long p = abs(dividend), q = abs(divisor);
        if(p < q)     //遞歸出口
            return 0;

        while(p - q >= 0){
            p -=  q;
            q = q + q;
            exp2++;
        }
        ret = divide(p, abs(divisor));    //figure out new_p
        ret += (0x1 << exp2) - 1;
        if(dividend < 0 && divisor > 0 || dividend > 0 && divisor < 0)
            ret = -ret;
        if(ret >= (unsigned int)(0x1<<31)) --ret;
        return ret;
    }
};