feat: add solutions to leetcode problem: No.0628

Author: yanglbme

solution/0600-0699/0628.Maximum Product of Three Numbers/README.md

@@ -39,15 +39,31 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def maximumProduct(self, nums: List[int]) -> int:
+        n = len(nums)
+        nums.sort()
+        # 全负 0 1 n-1
+        # 全正 n-1 n-2 n-3
+        # 有正有负 max([0 1 n-1], [n-1 n-2 n-3])
+        return max(nums[0] * nums[1] * nums[n - 1], nums[n - 1] * nums[n - 2] * nums[n - 3])
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int maximumProduct(int[] nums) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        // 全负 0 1 n-1
+        // 全正 n-1 n-2 n-3
+        // 有正有负 max([0 1 n-1], [n-1 n-2 n-3])
+        return Math.max(nums[0] * nums[1] * nums[n - 1], nums[n - 1] * nums[n - 2] * nums[n - 3]);
+    }
+}
 ```
 
 ### **...**

solution/0600-0699/0628.Maximum Product of Three Numbers/README_EN.md

@@ -46,13 +46,23 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def maximumProduct(self, nums: List[int]) -> int:
+        n = len(nums)
+        nums.sort()
+        return max(nums[0] * nums[1] * nums[n - 1], nums[n - 1] * nums[n - 2] * nums[n - 3])
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int maximumProduct(int[] nums) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        return Math.max(nums[0] * nums[1] * nums[n - 1], nums[n - 1] * nums[n - 2] * nums[n - 3]);
+    }
+}
 ```
 
 ### **...**

solution/0600-0699/0628.Maximum Product of Three Numbers/Solution.java

@@ -0,0 +1,7 @@
+class Solution {
+    public int maximumProduct(int[] nums) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        return Math.max(nums[0] * nums[1] * nums[n - 1], nums[n - 1] * nums[n - 2] * nums[n - 3]);
+    }
+}

solution/0600-0699/0628.Maximum Product of Three Numbers/Solution.py

@@ -0,0 +1,5 @@
+class Solution:
+    def maximumProduct(self, nums: List[int]) -> int:
+        n = len(nums)
+        nums.sort()
+        return max(nums[0] * nums[1] * nums[n - 1], nums[n - 1] * nums[n - 2] * nums[n - 3])