29. Divide Two Integers

最新推荐文章于 2025-08-22 15:10:27 发布

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最新推荐文章于 2025-08-22 15:10:27 发布

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文章标签：开发工具

原文链接：http://www.cnblogs.com/argenbarbie/p/5243847.html

本文介绍了一种不使用乘法、除法和模运算符来实现整数除法的方法，包括对数运算和二进制指数运算两种解决方案，并讨论了溢出处理。此外，提供了一个基于位操作的二进制指数树启发式解决方案。

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

If it is overflow, return MAX_INT.

(1) log

int divide(int dividend, int divisor) {
    if(dividend == 0)
        return 0;
    if(divisor == 0)
        return INT_MAX;
    double t1 = log(fabs(dividend)), t2 = log(fabs(divisor));
    long long ans = (double)exp(t1-t2);
    if((dividend > 0) ^ (divisor > 0))
        ans = -ans;
    if(ans > INT_MAX)
        return INT_MAX;
    return ans;
}

注意：

abs(-2147483648) = -2147483648, fabs(-2147483648) = 2147483648

(2) Binary Index tree idea inspired solution, as we can decompose any result number to sum of the power of 2.

int divide(int dividend, int divisor) {
    long long result = 0;
    long long m = abs((long long)dividend);
    long long n = abs((long long)divisor);
    while (m >= n) {
        long long s = n, power = 1;
        while ((s << 1) <= m)
        {
            s <<= 1;
            power <<= 1;
        }
        result += power;
        m -= s;
    }
    if ((dividend>0) ^ (divisor>0))
        result = -result;
    return result>INT_MAX ? INT_MAX : result;
}