diff --git a/leetcode/Flynnon/medium/29-191014-Divide_Two_Integers.py b/leetcode/Flynnon/medium/29-191014-Divide_Two_Integers.py
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+"""
+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: [−2^31,  2^31 − 1]. For the purpose of this problem, assume that your function returns 2^31 − 1 when the division result overflows.
+
+链接：https://leetcode-cn.com/problems/divide-two-integers
+"""
+
+
+class Solution:
+    def divide(self, dividend: int, divisor: int) -> int:
+        """
+            x // y == 2^x + 2^y + ... + 0
+            归根到底还是减法，只是每次尽可能多的减去值(考虑到不能用乘除法，直接用位移)，可以认为是贪心
+        """
+
+        assert divisor != 0
+
+        symbol = 1
+        if dividend < 0:
+            dividend = -dividend
+            symbol = -1
+
+        if divisor < 0:
+            divisor = -divisor
+            symbol = 1 if symbol == -1 else -1
+
+        multiple = 0
+        # 当被除数大于除数的时候才需要继续
+        while dividend >= divisor > 0:
+
+            # 本次的最大倍数(2^t_multiple <= dividend && 2^(t_multiple + 1) > dividend)
+            t_multiple = 1
+            tmp_divisor = divisor
+
+            # 求当前的最大2倍数
+            while True:
+                if tmp_divisor > dividend:
+                    break
+
+                # 简单的移位
+                tmp_divisor <<= 1
+                t_multiple <<= 1
+
+            # 上面会多移位一次，此处移回来
+            t_multiple >>= 1
+            dividend -= (tmp_divisor >> 1)
+
+            # 将目前的倍数增加至最大倍数中
+            multiple += t_multiple
+
+        # 这个不好说算不算乘法，但不这么写就写不了了....
+        if symbol == -1:
+            multiple = -multiple
+
+        max_value = 0x7fffffff
+        # 这里对结果进行裁剪，在python中无数字大小的限制，因此....
+        if multiple > max_value or multiple < -(max_value + 1):
+            return max_value
+        return multiple