solve problem License Key Formatting

Author: P-ppc

README.md

@@ -172,6 +172,7 @@ All solutions will be accepted!
 |160|[Intersection Of Two Linked Lists](https://leetcode-cn.com/problems/intersection-of-two-linked-lists/description/)|[java/py/js](./algorithms/IntersectionOfLinkedLists)|Easy|
 |860|[LemonadeChange](https://leetcode-cn.com/problems/lemonade-change/description/)|[java/py/js](./algorithms/LemonadeChange)|Easy|
 |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|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/LicenseKeyFormatting/README.md

@@ -0,0 +1,2 @@
+# License Key Formatting
+This problem is easy to solve

algorithms/LicenseKeyFormatting/Solution.java

@@ -0,0 +1,25 @@
+class Solution {
+    public String licenseKeyFormatting(String S, int K) {
+        S = S.toUpperCase();
+        
+        StringBuilder res = new StringBuilder();
+        int count = 0;
+        
+        for (int i = S.length() - 1; i >= 0; i--) {
+            char c = S.charAt(i);
+            
+            if (c != '-') {
+                if (count == K) {
+                    res.insert(0, '-');
+                    res.insert(0, c);
+                    count = 1;
+                } else if (count < K) {
+                    res.insert(0, c);
+                    count++;
+                }
+            }
+        }
+        
+        return res.toString();
+    }
+}

algorithms/LicenseKeyFormatting/solution.js

@@ -0,0 +1,25 @@
+/**
+ * @param {string} S
+ * @param {number} K
+ * @return {string}
+ */
+var licenseKeyFormatting = function(S, K) {
+    S = S.toUpperCase()
+    
+    let res = '',
+        count = 0
+    
+    for (let i = S.length - 1; i >= 0; i--) {
+        if (S[i] !== '-') {
+            if (count === K) {
+                res = S[i] + '-' + res
+                count = 1
+            } else if (count < K) {
+                res = S[i] + res
+                count++
+            }
+        }
+    }
+    
+    return res
+};

algorithms/LicenseKeyFormatting/solution.py

@@ -0,0 +1,22 @@
+class Solution(object):
+    def licenseKeyFormatting(self, S, K):
+        """
+        :type S: str
+        :type K: int
+        :rtype: str
+        """
+        S = S.upper()
+        res = ''
+        
+        count = -1
+        
+        length = len(S)
+        for i in range(length):
+            if S[length - 1 - i] != '-':
+                if count == K - 1:
+                    res = S[length - 1 - i] + '-' + res
+                    count = 0
+                elif count < K - 1:
+                    res = S[length - 1 - i] + res
+                    count += 1
+        return res