LeetCode题解：Divide Two Integers-优快云博客

本文链接：https://blog.youkuaiyun.com/MagiSu/article/details/17017957

本文介绍了一种不使用乘法、除法及取余运算符实现两个整数相除的方法。核心思路是通过不断将试除数加倍来逼近被除数，同时记录商的变化。此外，文章还考虑了整数溢出等特殊情况。

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Divide Two Integers

Divide two integers without using multiplication, division and mod operator.

思路：

核心思想是：令试除数等于除数，如果试除数小于被除数，那么试除数乘二，商乘二，继续比较；否则试除数除二，商除二，被除数减去试除数，重新开始。这其实就是类似二分搜索法。

题目有一些边边角角的特殊情况，比如int的最大值比最小值的绝对值小1之类的。所以比较繁琐。

题解：

class Solution {
public:
int divide (int dividend, int divisor)
{
//    assert (divisor != 0);
    
    if (divisor == 1)
        return dividend;
    else if (divisor == -1)
        return -dividend;
    
    // verify the sign of the result
    const int flag_sgn = (1 << (sizeof (int) * 8 - 1));
    bool sgn = true;    // positive
    if ( (dividend & flag_sgn) ^ (divisor & flag_sgn))
        sgn = false;    // negative
        
    // check if the negative number is numeric_limits<int>::min()
    // this requires special care, as abs(INT_MIN) = INT_MAX + 1
    if (dividend == numeric_limits<int>::min())
    {
        switch (divisor)
        {
        case 1:
            return dividend;
        case -1:
            // problematic
            return -dividend;
        case 2:
            return - (numeric_limits<int>::max() >> 1) - 1;
        case -2:
            return (numeric_limits<int>::max() >> 1) + 1;
        case numeric_limits<int>::min():
            return 1;
        default:
            dividend = numeric_limits<int>::max();
        }
    }
    
    if (dividend < 0) dividend = - dividend;
    if (divisor < 0) divisor = - divisor;
    
    // now do the real job
    
    while (! ( (divisor) & 1))
        divisor >>= 1, dividend >>= 1;
        
    int result = 0;
    while (dividend > 0)
    {
        int trial = divisor;
        int accumu = 1;
        bool overflow = false;
        while (trial <= dividend)
        {
            if (trial > (numeric_limits<int>::max() >> 1))
            {
                overflow = true;
                break;
            }
            trial <<= 1, accumu <<= 1;
        }
        
        if (!overflow) trial >>= 1, accumu >>= 1;
        dividend -= trial;
        result += accumu;
    }
    
    if (!sgn) 
        result = - result;
    
    return result;
}
};