solve problem Rotate String

Author: P-ppc

README.md

@@ -119,6 +119,7 @@ All solutions will be accepted!
 |203|[Remove Linked List Elements](https://leetcode-cn.com/problems/remove-linked-list-elements/description/)|[java/py/js](./algorithms/RemoveLinkedListElements)|Easy|
 |237|[Delete Node In A Linked List](https://leetcode-cn.com/problems/delete-node-in-a-linked-list/description/)|[java/py/js](./algorithms/DeleteNodeInALinkedList)|Easy|
 |405|[Convert A Number To Hexadecimal](https://leetcode-cn.com/problems/convert-a-number-to-hexadecimal/description/)|[java/py/js](./algorithms/ConvertANumberToHexadecimal)|Easy|
+|796|[Rotate String](https://leetcode-cn.com/problems/rotate-string/description/)|[java/py/js](./algorithms/RotateString)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/RotateString/README.md

@@ -0,0 +1,2 @@
+# Rotate String
+This problem is easy to solve

algorithms/RotateString/Solution.java

@@ -0,0 +1,14 @@
+class Solution {
+    public boolean rotateString(String A, String B) {
+        if (A.length() != B.length()) return false;
+        
+        int i = 0;
+        while (i <= A.length()) {
+            if (A.equals(B)) return true;
+            A = A.substring(1) + A.substring(0, 1);
+            i++;
+        }
+        
+        return false;
+    }
+}

algorithms/RotateString/solution.js

@@ -0,0 +1,17 @@
+/**
+ * @param {string} A
+ * @param {string} B
+ * @return {boolean}
+ */
+var rotateString = function(A, B) {
+    if (A.length !== B.length) return false
+    
+    let i = 0
+    while (i <= A.length) {
+        if (A === B) return true
+        A = A.substr(1) + A.substr(0, 1)
+        i++
+    }
+    
+    return false
+};

algorithms/RotateString/solution.py

@@ -0,0 +1,18 @@
+class Solution(object):
+    def rotateString(self, A, B):
+        """
+        :type A: str
+        :type B: str
+        :rtype: bool
+        """
+        if len(A) != len(B):
+            return False
+        
+        i = 0
+        while i <= len(A):
+            if A == B:
+                return True
+            A = A[1:] + A[0]
+            i += 1
+            
+        return False