29. Divide Two Integers

Problem

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3

Example 2:

Input: dividend = 7, divisor = -3
Output: -2

Note:

• Both dividend and divisor will be 32-bit signed integers.
• The divisor will never be 0.
• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.

Solution

/**
 * @param {number} dividend
 * @param {number} divisor
 * @return {number}
 */
var divide = function(dividend, divisor) {
  var did = Math.abs(dividend);
  var dis = Math.abs(divisor);
  var sign = (divisor > 0 && dividend > 0) || (divisor < 0 && dividend < 0);
  var res = 0;
  var arr = [dis];

  if (dividend === 0 || did < dis) return 0;
  if (divisor === -1 && dividend === -2147483648) return 2147483647;
  if (dis === 1) return divisor > 0 ? dividend : -dividend;

  while (arr[arr.length - 1] < did) arr.push(arr[arr.length - 1] + arr[arr.length - 1]);

  for (var i = arr.length - 1; i >= 0; i--) {
    if (did >= arr[i]) {
      did -= arr[i];
      res += i === 0 ? 1 : 2 << (i - 1);
    }
  }

  return sign ? res : -res;
};

Explain:

1. 要注意 -2147483648 / -1 越界的情况
2. 核心就是 dividend -= divisor << i; result += 2 << (i - 1);
3. 其它语言有 long、long long 类型可以保证 divisor << i 不越界，但是 js 没有。比如 2 << 29 是 1073741824，2 << 30 就越界了，不会得到预期的结果。我这里是用加法提前计算好 divisor << i。

Complexity:

• Time complexity : O(log(n)).
• Space complexity : O(log(n)).