題目地址：

中文：https://leetcode-cn.com/problems/divide-two-integers/
英文：https://leetcode.com/problems/divide-two-integers/

題目描述：

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, which means losing its fractional part. For example, truncate(8.345) = 8 and truncate(-2.7335) = -2.

Note:

• Assume we are dealing with an environment that could only store
  integers within the 32-bit signed integer range: [−2^31, 2^31 − 1].
  For this problem, assume that your function returns 2^31 − 1 when the
  division result overflows.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = truncate(3.33333..) = 3.

Example 2:

Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = truncate(-2.33333..) = -2.

Example 3:

Input: dividend = 0, divisor = 1
Output: 0

Example 4:

Input: dividend = 1, divisor = 1
Output: 1

Constraints:

-2^31 <= dividend, divisor <= 2^31 - 1
divisor != 0

思路：

用被除數迴圈去減除數，會超時。
換了個思路，除數每次增加兩倍，直到大於被除數，再減掉多餘的部分。
有點計網擁塞控制的視窗指數增長的意思。

題解：

class Solution {
    public static int divide(int dividend, int divisor) {
        long res=1;
        int flag_dividend = 1;
        int flag_divisor = 1;
        long d1 = dividend;
        long d2 = divisor;
        if(d1<0) {flag_dividend = -1;d1=-d1;}
        if(d2<0) {flag_divisor = -1;d2=-d2;}
        if(d1<d2) return 0;
        while(d1>d2){
            d2 *= 2;
            res*= 2;
        }
        while (d2>d1){
            d2=d2-Math.abs(divisor);
            res--;
        }
        res=res*flag_dividend*flag_divisor;
        if(res>Integer.MAX_VALUE||res<Integer.MIN_VALUE) return Integer.MAX_VALUE;
        return (int)res;
    }
}