solve problem Rectangle Overlap

Author: P-ppc

README.md

@@ -174,6 +174,7 @@ All solutions will be accepted!
 |401|[Binary Watch](https://leetcode-cn.com/problems/binary-watch/description/)|[java/py/js](./algorithms/BinaryWatch)|Easy|
 |482|[License Key Formatting](https://leetcode-cn.com/problems/license-key-formatting/description/)|[java/py/js](./algorithms/LicenseKeyFormatting)|Easy|
 |720|[Longest Word In Dictionary](https://leetcode-cn.com/problems/longest-word-in-dictionary/description/)|[java/py/js](./algorithms/LongestWordInDictionary)|Easy|
+|836|[Rectangle Overlap](https://leetcode-cn.com/problems/rectangle-overlap/description/)|[java/py/js](./algorithms/RectangleOverlap)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/RectangleOverlap/README.md

@@ -0,0 +1,2 @@
+# Rectangle Overlap
+This problem is easy to solve

algorithms/RectangleOverlap/Solution.java

@@ -0,0 +1,11 @@
+class Solution {
+    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
+        if (rec1[0] >= rec2[2] || rec2[0] >= rec1[2]) {
+            return false;
+        } else if (rec1[1] >= rec2[3] || rec2[1] >= rec1[3]) {
+            return false;
+        }
+
+        return true;
+    }
+}

algorithms/RectangleOverlap/solution.js

@@ -0,0 +1,14 @@
+/**
+ * @param {number[]} rec1
+ * @param {number[]} rec2
+ * @return {boolean}
+ */
+var isRectangleOverlap = function(rec1, rec2) {
+    if (rec1[0] >= rec2[2] || rec2[0] >= rec1[2]) {
+        return false
+    } else if (rec1[1] >= rec2[3] || rec2[1] >= rec1[3]) {
+        return false
+    }
+    
+    return true
+};

algorithms/RectangleOverlap/solution.py

@@ -0,0 +1,13 @@
+class Solution(object):
+    def isRectangleOverlap(self, rec1, rec2):
+        """
+        :type rec1: List[int]
+        :type rec2: List[int]
+        :rtype: bool
+        """
+        if rec1[0] >= rec2[2] or rec2[0] >= rec1[2]:
+            return False
+        elif rec1[1] >= rec2[3] or rec2[1] >= rec1[3]:
+            return False
+        return True
+