Java LeetCode除以两个整数-重复指数搜索负数解

我很难理解leetcode中下面的解决方案。为什么int powerOfTwo=-1用-1初始化，因为我们已经处理了divide（int_MIN，-1）的情况 : 添加问题陈述-

给定两个被除数和除数的整数，不使用乘法、除法和mod运算符对两个整数进行除法

返回除数除以除数后的商

整数除法应该向零截断，这意味着丢失其小数部分。例如，truncate（8.345）=8，truncate（-2.7335）=-2

注:

假设我们处理的环境只能存储32位有符号整数范围内的整数：[−2^31, 2^31 − 1]. 对于这个问题，假设函数返回2^31− 1当除法结果溢出时

例1：

输入：除数=10，除数=3 产出：3 说明：10/3=截断（3.33333..）=3

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还请解释代码所针对的问题。您的链接只适用于leetcode的高级用户。@ManuelRenePaulsToews当然可以。在原始帖子中添加。谢谢，请解释一下代码所针对的问题。您的链接只适用于leetcode的高级用户。@ManuelRenePaulsToews当然可以。在原始帖子中添加。谢谢

private static int HALF_INT_MIN = -1073741824;

public int divide(int dividend, int divisor) {
    // Special case: overflow.
    if (dividend == Integer.MIN_VALUE && divisor == -1) {
        return Integer.MAX_VALUE;
    }

    /* We need to convert both numbers to negatives.
     * Also, we count the number of negatives signs. */
    int negatives = 2;
    if (dividend > 0) {
        negatives--;
        dividend = -dividend;
    }
    if (divisor > 0) {
        negatives--;
        divisor = -divisor;
    }

    int quotient = 0;
    /* Once the divisor is bigger than the current dividend,
     * we can't fit any more copies of the divisor into it. */
    while (divisor >= dividend) {
        /* We know it'll fit at least once as divivend >= divisor.
         * Note: We use a negative powerOfTwo as it's possible we might have
         * the case divide(INT_MIN, -1). */
        int powerOfTwo = -1;
        int value = divisor;
        /* Check if double the current value is too big. If not, continue doubling.
         * If it is too big, stop doubling and continue with the next step */
        while (value >= HALF_INT_MIN && value + value >= dividend) {
            value += value;
            powerOfTwo += powerOfTwo;
        }
        // We have been able to subtract divisor another powerOfTwo times.
        quotient += powerOfTwo;
        // Remove value so far so that we can continue the process with remainder.
        dividend -= value;
    }

    /* If there was originally one negative sign, then
     * the quotient remains negative. Otherwise, switch
     * it to positive. */
    if (negatives != 1) {
        return -quotient;
    }
    return quotient;
}