LeetCode T29 Divide Two Integers

题目地址：

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

题目描述：

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：

用被除数循环去减除数，会超时。
换了个思路，除数每次增加两倍，直到大于被除数，再减掉多余的部分。
有点计网拥塞控制的窗口指数增长的意思。

题解：

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;
    }
}