feat: add python and java solutions to leetcode problem: No.0152

Author: yanglbme

README.md

@@ -132,6 +132,7 @@
 ### 动态规划
 
 - [最大子序和](/solution/0000-0099/0053.Maximum%20Subarray/README.md)
+- [乘积最大子序列](/solution/0100-0199/0152.Maximum%20Product%20Subarray/README.md)
 - [打家劫舍](/solution/0100-0199/0198.House%20Robber/README.md)
 - [打家劫舍 II](/solution/0200-0299/0213.House%20Robber%20II/README.md)
 - [最长上升子序列](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README.md)

README_EN.md

@@ -121,6 +121,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 ### Dynamic Programming
 
 - [Maximum Subarray](/solution/0000-0099/0053.Maximum%20Subarray/README_EN.md)
+- [Maximum Product Subarray](/solution/0100-0199/0152.Maximum%20Product%20Subarray/README_EN.md)
 - [House Robber](/solution/0100-0199/0198.House%20Robber/README_EN.md)
 - [House Robber II](/solution/0200-0299/0213.House%20Robber%20II/README_EN.md)
 - [Longest Increasing Subsequence](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README_EN.md)

solution/0100-0199/0152.Maximum Product Subarray/README.md

@@ -24,22 +24,54 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+考虑当前位置 i：
+
+- 如果是一个负数的话，那么我们希望以它前一个位置结尾的某个段的积也是个负数，这样可以负负得正，并且我们希望这个积尽可能「负得更多」，即尽可能小。
+- 如果是一个正数的话，我们更希望以它前一个位置结尾的某个段的积也是个正数，并且希望它尽可能地大。
+
+因此，分别维护 fmax 和 fmin。
+
+- `fmax(i) = max(nums[i], fmax(i - 1) * nums[i], fmin(i - 1) * nums[i])`
+- `fmin(i) = min(nums[i], fmax(i - 1) * nums[i], fmin(i - 1) * nums[i])`
+- `res = max(fmax(i)), i∈[0, n)`
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def maxProduct(self, nums: List[int]) -> int:
+        maxf = minf = nums[0]
+        res, n = nums[0], len(nums)
+        for i in range(1, n):
+            p, q = maxf, minf
+            maxf = max(nums[i], p * nums[i], q * nums[i])
+            minf = min(nums[i], p * nums[i], q * nums[i])
+            res = max(res, maxf)
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int maxProduct(int[] nums) {
+        int maxf = nums[0], minf = nums[0];
+        int res = nums[0], n = nums.length;
+        for (int i = 1; i < n; ++i) {
+            int p = maxf, q = minf;
+            maxf = Math.max(nums[i], Math.max(p * nums[i], q * nums[i]));
+            minf = Math.min(nums[i], Math.min(p * nums[i], q * nums[i]));
+            res = Math.max(res, maxf);
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0100-0199/0152.Maximum Product Subarray/README_EN.md

@@ -35,13 +35,34 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def maxProduct(self, nums: List[int]) -> int:
+        maxf = minf = nums[0]
+        res, n = nums[0], len(nums)
+        for i in range(1, n):
+            p, q = maxf, minf
+            maxf = max(nums[i], p * nums[i], q * nums[i])
+            minf = min(nums[i], p * nums[i], q * nums[i])
+            res = max(res, maxf)
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int maxProduct(int[] nums) {
+        int maxf = nums[0], minf = nums[0];
+        int res = nums[0], n = nums.length;
+        for (int i = 1; i < n; ++i) {
+            int p = maxf, q = minf;
+            maxf = Math.max(nums[i], Math.max(p * nums[i], q * nums[i]));
+            minf = Math.min(nums[i], Math.min(p * nums[i], q * nums[i]));
+            res = Math.max(res, maxf);
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0100-0199/0152.Maximum Product Subarray/Solution.java

@@ -1,14 +1,12 @@
 class Solution {
     public int maxProduct(int[] nums) {
-        int max[] = new int[nums.length];
-        int min[] = new int[nums.length];
-        int res = nums[0];
-        max[0] =  min[0] = nums[0];
-        for(int i=1;i<nums.length;i++){
-            int num = nums[i];
-            max[i] = Math.max(Math.max(max[i-1]*num,min[i-1]*num),num);
-            min[i] = Math.min(Math.min(min[i-1]*num,max[i-1]*num),num);
-            res = Math.max(max[i],res);
+        int maxf = nums[0], minf = nums[0];
+        int res = nums[0], n = nums.length;
+        for (int i = 1; i < n; ++i) {
+            int p = maxf, q = minf;
+            maxf = Math.max(nums[i], Math.max(p * nums[i], q * nums[i]));
+            minf = Math.min(nums[i], Math.min(p * nums[i], q * nums[i]));
+            res = Math.max(res, maxf);
         }
         return res;
     }

solution/0100-0199/0152.Maximum Product Subarray/Solution.py

@@ -1,21 +1,10 @@
-
-# Beats 88% in python.
-class Solution(object):
-    def maxProduct(self, nums):
-        """
-        :type nums: List[int]
-        :rtype: int
-        """
-        if not nums:
-            return 0
-        localMin = localMax = maxi = nums[0]
-        
-        for i in range(1, len(nums)):
-            if nums[i] < 0:
-                localMin, localMax = localMax, localMin
-            
-            localMin = min(nums[i], localMin * nums[i])
-            localMax = max(nums[i], localMax * nums[i])
-            maxi = max(localMax, maxi)
-            
-        return maxi
+class Solution:
+    def maxProduct(self, nums: List[int]) -> int:
+        maxf = minf = nums[0]
+        res, n = nums[0], len(nums)
+        for i in range(1, n):
+            p, q = maxf, minf
+            maxf = max(nums[i], p * nums[i], q * nums[i])
+            minf = min(nums[i], p * nums[i], q * nums[i])
+            res = max(res, maxf)
+        return res