给定两个整数，被除数 dividend 和除数 divisor。将两数相除，要求不使用乘法、除法和 mod 运算符。返回被除数 dividend 除以除数 divisor 得到的商。整数除法的结果应当截去（truncate）其小数部分，例如：truncate(8.345) = 8 以及 truncate(-2.7335) = -2 示例 1:输入: dividend = 10, divisor = 3输出: 3解释: 10/3 = truncate(3.33333..) = truncate(3) = 3示例 2:输入: dividend = 7, divisor = -3输出: -2解释: 7/-3 = truncate(-2.33333..) = -2 提示：被除数和除数均为 32 位有符号整数。除数不为 0。假设我们的环境只能存储 32 位有符号整数，其数值范围是 [−231,  231 − 1]。本题中，如果除法结果溢出，则返回 231 − 1。

let acc = divisor;let count = 0;while (dividend - acc >= 0) {  acc += divisor;  count++;}return count;

const isNegative = dividend > 0 !== divisor > 0;

const isNegative = dividend ^ (divisor < 0);

/* * @lc app=leetcode id=29 lang=javascript * * [29] Divide Two Integers *//** * @param {number} dividend * @param {number} divisor * @return {number} */var divide = function (dividend, divisor) {  if (divisor === 1) return dividend;  // 这种方法很巧妙，即符号相同则为正，不同则为负  const isNegative = dividend > 0 !== divisor > 0;  const MAX_INTERGER = Math.pow(2, 31);  const res = helper(Math.abs(dividend), Math.abs(divisor));  // overflow  if (res > MAX_INTERGER - 1 || res < -1 * MAX_INTERGER) {    return MAX_INTERGER - 1;  }  return isNegative ? -1 * res : res;};function helper(dividend, divisor) {  // 二分法  if (dividend <= 0) return 0;  if (dividend < divisor) return 0;  if (divisor === 1) return dividend;  let acc = 2 * divisor;  let count = 1;  while (dividend - acc > 0) {    acc += acc;    count += count;  }  // 直接使用位移运算，比如acc >> 1会有问题  const last = dividend - Math.floor(acc / 2);  return count + helper(last, divisor);}

class Solution:    def divide(self, dividend: int, divisor: int) -> int:        """        二分法        :param int divisor        :param int dividend        :return int        """        # 错误处理        if divisor == 0:            raise ZeroDivisionError        if abs(divisor) == 1:            result = dividend if 1 == divisor else -dividend            return min(2**31-1, max(-2**31, result))        # 确定结果的符号        sign = ((dividend >= 0) == (divisor >= 0))        result = 0        # abs也可以直接写在while条件中，不过可能会多计算几次        _divisor = abs(divisor)        _dividend = abs(dividend)        while _divisor <= _dividend:            r, _dividend = self._multi_divide(_divisor, _dividend)            result += r        result = result if sign else -result        # 注意返回值不能超过32位有符号数的表示范围        return min(2**31-1, max(-2**31, result))    def _multi_divide(self, divisor, dividend):        """        翻倍除法，如果可以被除，则下一步除数翻倍，直至除数大于被除数，        返回商加总的结果与被除数的剩余值；        这里就不做异常处理了；        :param int divisor        :param int dividend        :return tuple result, left_dividend        """        result = 0        times_count = 1        while divisor <= dividend:            dividend -= divisor            result += times_count            times_count += times_count            divisor += divisor        return result, dividend

class Solution {public:    int divide(int dividend, int divisor) {        if (!divisor) return 0;  // divide-by-zero error        bool pos1 = dividend > 0, pos2 = divisor > 0, pos = !(pos1^pos2);        if (pos1) dividend = -dividend;        if (pos2) divisor = -divisor;        int q = 0, d = divisor, t = 1;        while (t > 0 && dividend < 0) {            if (dividend - d <= 0) {                dividend -= d;                q -= t;                if ((INT_MIN >> 1) < d) {                    t <<= 1;                    d <<= 1;                }            } else {                d >>= 1;                t >>= 1;            }        }        return pos? -q : q;    }};