LeetCode Java First 400 题解-029_divisor first sum-优快云博客

本文链接：https://blog.youkuaiyun.com/felomeng/article/details/102850400

本文介绍了一种在不使用乘法、除法和取模运算的情况下实现两个整数相除的方法。通过模拟除法过程，每次将被除数加倍，相应的商也加倍，以此来找到最大的被除数倍数不超过除数的情况，从而计算出商。这种方法的时间复杂度为O((lgn)^2)。

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

If it is overflow, return MAX_INT.

public int divide(int dividend, int divisor) {

    if (divisor == 0)

        throw new java.lang.ArithmeticException("/ by zero");

    long result = divideLong(dividend, divisor);

    return result > Integer.MAX_VALUE ? Integer.MAX_VALUE : (int) result;

}



public long divideLong(long dividend, long divisor) {

    boolean negative = dividend < 0 != divisor < 0;

    if (dividend < 0)

        dividend = -dividend;

    if (divisor < 0)

        divisor = -divisor;

    if (dividend < divisor)

        return 0;

    long sum = divisor;

    long divide = 1;

    while ((sum + sum) <= dividend) {

        sum += sum;

        divide += divide;

    }

    return negative ? -(divide + divideLong((dividend - sum), divisor))

            : (divide + divideLong((dividend - sum), divisor));

}