add the solution "820 - Short Encoding of Words" into readme, and add the comments, adjust the filename and put it into its directory haoel#216

Author: haoel

README.md

@@ -54,6 +54,7 @@ LeetCode
 |844|[Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/description/) | [C++](./algorithms/cpp/backspaceStringCompare/BackspaceStringCompare.cpp)|Easy|
 |837|[Most Common Word](https://leetcode.com/problems/most-common-word/) | [C++](./algorithms/cpp/mostCommonWord/MostCommonWord.cpp)|Easy|
 |830|[Positions of Large Groups](https://leetcode.com/problems/positions-of-large-groups/) | [Python](./algorithms/python/PositionsOfLargeGroups/largeGroupPositions.py)|Easy|
+|820|[Short Encoding of Words](https://leetcode.com/problems/short-encoding-of-words/) | [C++](./algorithms/cpp/shortEncodingOfWords/ShortEncodingOfWords.cpp)|Medium|
 |804|[Unique Morse Code Words](https://leetcode.com/problems/unique-morse-code-words/description/) | [C++](./algorithms/cpp/uniqueMorseCodeWords/UniqueMorseCodeWords.cpp)|Easy|
 |771|[Jewels and Stones](https://leetcode.com/problems/jewels-and-stones/description) | [C++](./algorithms/cpp/jewelsAndStones/JewelsAndStones.cpp)|Easy|
 |747|[Largest Number At Least Twice of Others](https://leetcode.com/problems/largest-number-at-least-twice-of-others/) | [Python](./algorithms/python/LargestNumberAtLeastTwiceOfOthers/dominantIndex.py)|Easy|

algorithms/cpp/Short_Encoding_of_Words.cpp

@@ -0,0 +1,57 @@
+// Source : https://leetcode.com/problems/short-encoding-of-words/
+// Author : Hao Chen
+// Date   : 2020-10-02
+/***************************************************************************************************** 
+ *
+ * Given a list of words, we may encode it by writing a reference string S and a list of indexes A.
+ * 
+ * For example, if the list of words is ["time", "me", "bell"], we can write it as S = "time#bell#" 
+ * and indexes = [0, 2, 5].
+ * 
+ * Then for each index, we will recover the word by reading from the reference string from that index 
+ * until we reach a "#" character.
+ * 
+ * What is the length of the shortest reference string S possible that encodes the given words?
+ * 
+ * Example:
+ * 
+ * Input: words = ["time", "me", "bell"]
+ * Output: 10
+ * Explanation: S = "time#bell#" and indexes = [0, 2, 5].
+ * 
+ * Note:
+ * 
+ * 	1 <= words.length <= 2000.
+ * 	1 <= words[i].length <= 7.
+ * 	Each word has only lowercase letters.
+ * 
+ ******************************************************************************************************/
+class Solution {
+public:
+    static bool comp(string a,string b){
+        return a.size() < b.size();
+    }
+    int minimumLengthEncoding(vector<string>& words) {
+        sort(words.begin(),words.end(),comp);
+        int ans = 0;
+        int count = 0;
+        unordered_map<string,int> M;
+        for(int i = 0 ; i < words.size() ; i++){
+            string temp = "";
+            for(int k = words[i].size() - 1 ; k >= 0 ; k--){
+                temp = words[i][k] + temp;
+                M[temp]++;
+            }
+        }
+        for(int i = 0 ; i < words.size() ; i++){
+            ans = ans + words[i].size();
+            count++;
+            if(M[words[i]] > 1){
+                ans = ans - words[i].size();
+                count--;
+                M[words[i]]--;
+            }
+        }
+        return ans + count;
+    }
+};