solve problem Maximum Product Of Three Numbers

Author: P-ppc

README.md

@@ -141,6 +141,7 @@ All solutions will be accepted!
 |125|[Valid Palindrome](https://leetcode-cn.com/problems/valid-palindrome/description/)|[java/py/js](./algorithms/ValidPalindrome)|Easy|
 |141|[Linked List Cycle](https://leetcode-cn.com/problems/linked-list-cycle/description/)|[java/py/js](./algorithms/LinkedListCycle)|Easy|
 |387|[First Unique Character In A String](https://leetcode-cn.com/problems/first-unique-character-in-a-string/description/)|[java/py/js](./algorithms/FirstUniqueCharacterInAString)|Easy|
+|628|[Maximum Product Of Three Numbers](https://leetcode-cn.com/problems/maximum-product-of-three-numbers/description/)|[java/py/js](./algorithms/MaximumProductOfThreeNumbers)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/MaximumProductOfThreeNumbers/README.md

@@ -0,0 +1,2 @@
+# Maximum Product Of Three Numbers
+This problem is easy to solve

algorithms/MaximumProductOfThreeNumbers/Solution.java

@@ -0,0 +1,31 @@
+class Solution {
+    public int maximumProduct(int[] nums) {
+        int max1 = Integer.MIN_VALUE,
+            max2 = Integer.MIN_VALUE,
+            max3 = Integer.MIN_VALUE,
+            min1 = Integer.MAX_VALUE,
+            min2 = Integer.MAX_VALUE;
+        
+        for (int num : nums) {
+            if (num >= max1) {
+                max3 = max2;
+                max2 = max1;
+                max1 = num;
+            } else if (num >= max2) {
+                max3 = max2;
+                max2 = num;
+            } else if (num >= max3) {
+                max3 = num;
+            }
+
+            if (num <= min1) {
+                min2 = min1;
+                min1 = num;
+            } else if (num <= min2) {
+                min2 = num;
+            }
+        }
+        
+        return Math.max(max1 * max2 * max3, max1 * min1 * min2);
+    }
+}

algorithms/MaximumProductOfThreeNumbers/solution.js

@@ -0,0 +1,33 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maximumProduct = function(nums) {
+    let max1 = Number.MIN_SAFE_INTEGER,
+        max2 = Number.MIN_SAFE_INTEGER,
+        max3 = Number.MIN_SAFE_INTEGER,
+        min1 = Number.MAX_SAFE_INTEGER,
+        min2 = Number.MAX_SAFE_INTEGER
+    
+    nums.forEach(num => {
+        if (num >= max1) {
+            max3 = max2
+            max2 = max1
+            max1 = num
+        } else if (num >= max2) {
+            max3 = max2
+            max2 = num
+        } else if (num >= max3) {
+            max3 = num
+        }
+        
+        if (num <= min1) {
+            min2 = min1
+            min1 = num
+        } else if (num <= min2) {
+            min2 = num
+        }
+    })
+    
+    return Math.max(max1 * max2 * max3, max1 * min1 * min2)
+};

algorithms/MaximumProductOfThreeNumbers/solution.py

@@ -0,0 +1,33 @@
+class Solution(object):
+    def maximumProduct(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        # maximum product must be max1 * max2 * max3 or max1 * min1 * min2
+        
+        max1 = - sys.maxint
+        max2 = - sys.maxint
+        max3 = - sys.maxint
+        
+        min1 = sys.maxint
+        min2 = sys.maxint
+        
+        for num in nums:
+            if num >= max1:
+                max3 = max2
+                max2 = max1
+                max1 = num
+            elif num >= max2:
+                max3 = max2
+                max2 = num
+            elif num >= max3:
+                max3 = num
+            
+            if num <= min1:
+                min2 = min1
+                min1 = num
+            elif num <= min2:
+                min2 = num
+        
+        return max(max1 * max2 * max3, max1 * min1 * min2)