LeetCode29 Divide Two Integers

题目描述：
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Example 2:

Input: dividend = 7, divisor = -3
Output: -2
Note:

Both dividend and divisor will be 32-bit signed integers.
The divisor will never be 0.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 2^31 − 1 when the division result overflows.
题源：here；完整代码：here
思路：
两种方案：1 采用加法，让除数一直累加到被除数；2 采用位移，我们知道二进制数左移一位就相当于*2，我们让除数一直乘以二直到小于被除数；然后从大到小记录被除数一共包含了多少除数的倍数。
方案1代码 （提交超时）

    int divide(int dividend, int divisor) {
        int flag;
        double dividend_d = dividend, divisor_d = divisor;
        if (dividend_d >= 0 && divisor_d >= 0) flag = 1;
        else if (dividend_d >= 0 && divisor_d < 0) flag = -1, divisor_d = -divisor_d;
        else if (dividend_d < 0 && divisor_d>0) flag = -1, dividend_d = -dividend_d;
        else flag = 1, dividend_d = -dividend_d, divisor_d = -divisor_d;

        double result = 0;
        double sum = divisor_d;
        while (dividend_d >= sum){
            sum += divisor_d;
            result += flag;
        }
        if (result >= INT_MAX) return INT_MAX;
        if (result <= INT_MIN) return INT_MIN;
        return int(result);
    }

方案2代码

    int divide_2(int dividend, int divisor) {
        bool flag;
        long long dvd_abs, dvs_abs;
        if (dividend >= 0 && divisor >= 0) flag = true;
        else if (dividend >= 0 && divisor < 0) flag = false;
        else if (dividend < 0 && divisor >= 0) flag = false;
        else flag = true;

        unsigned int temp1 = dvd_abs = (dividend >= 0 ? dividend : -dividend);
        unsigned int temp2 = dvs_abs = divisor >= 0 ? divisor : -divisor;
        dvd_abs = temp1;
        dvs_abs = temp2;

        vector<unsigned int> value_record;
        unsigned long long value = 1;
        while (dvd_abs >= dvs_abs){
            value_record.push_back(dvs_abs);
            dvs_abs = dvs_abs << 1;
        }

        long long result = 0;
        unsigned int one = 1;
        for (int i = value_record.size() - 1; i >= 0; i--){
            if (dvd_abs >= value_record[i]) {
                result += one << i;
                dvd_abs -= value_record[i];
            }
        }

        if (!flag) result = 0 - result;

        if (result >= INT_MAX) return INT_MAX;
        else if (result <= INT_MIN) return INT_MIN;
        else return int(result);
    }