@@ -54,22 +54,160 @@ words[2] = "bell" ，s 开始于 indices[2] = 5 到下一个 '#
5454
5555<!-- 这里可写通用的实现逻辑 -->
5656
57+ 题目大意：充分利用重叠的后缀，使有效编码尽可能短
58+
5759<!-- tabs:start -->
5860
5961### ** Python3**
6062
6163<!-- 这里可写当前语言的特殊实现逻辑 -->
6264
6365``` python
64-
66+ class Trie :
67+ def __init__ (self None :
68+ self .children = [None ] * 26
69+
70+
71+ class Solution :
72+ def minimumLengthEncoding (self words : List[str ]) -> int :
73+ root = Trie()
74+ for w in words:
75+ cur = root
76+ for i in range (len (w) - 1 , - 1 , - 1 ):
77+ idx = ord (w[i]) - ord (' a'
78+ if cur.children[idx] == None :
79+ cur.children[idx] = Trie()
80+ cur = cur.children[idx]
81+ return self .dfs(root, 1 )
82+
83+ def dfs (self cur : Trie, l : int ) -> int :
84+ isLeaf, ans = True , 0
85+ for i in range (26 ):
86+ if cur.children[i] != None :
87+ isLeaf = False
88+ ans += self .dfs(cur.children[i], l + 1 )
89+ if isLeaf:
90+ ans += l
91+ return ans
6592```
6693
6794### ** Java**
6895
6996<!-- 这里可写当前语言的特殊实现逻辑 -->
7097
7198``` java
99+ class Trie {
100+ Trie [] children = new Trie [26 ];
101+ }
102+
103+ class Solution {
104+ public int minimumLengthEncoding (String [] words ) {
105+ Trie root = new Trie ();
106+ for (String w : words) {
107+ Trie cur = root;
108+ for (int i = w. length() - 1 ; i >= 0 ; i-- ) {
109+ int idx = w. charAt(i) - ' a'
110+ if (cur. children[idx] == null ) {
111+ cur. children[idx] = new Trie ();
112+ }
113+ cur = cur. children[idx];
114+ }
115+ }
116+ return dfs(root, 1 );
117+ }
118+
119+ private int dfs (Trie cur , int l ) {
120+ boolean isLeaf = true ;
121+ int ans = 0 ;
122+ for (int i = 0 ; i < 26 ; i++ ) {
123+ if (cur. children[i] != null ) {
124+ isLeaf = false ;
125+ ans += dfs(cur. children[i], l + 1 );
126+ }
127+ }
128+ if (isLeaf) {
129+ ans += l;
130+ }
131+ return ans;
132+ }
133+ }
134+ ```
135+
136+ ### ** Go**
137+
138+ ``` go
139+ type trie struct {
140+ children [26 ]*trie
141+ }
142+
143+ func minimumLengthEncoding (words []string ) int {
144+ root := new (trie)
145+ for _ , w := range words {
146+ cur := root
147+ for i := len (w) - 1 ; i >= 0 ; i-- {
148+ if cur.children [w[i]-' a' nil {
149+ cur.children [w[i]-' a' new (trie)
150+ }
151+ cur = cur.children [w[i]-' a'
152+ }
153+ }
154+ return dfs (root, 1 )
155+ }
156+
157+ func dfs (cur *trie , l int ) int {
158+ isLeaf , ans := true , 0
159+ for i := 0 ; i < 26 ; i++ {
160+ if cur.children [i] != nil {
161+ isLeaf = false
162+ ans += dfs (cur.children [i], l+1 )
163+ }
164+ }
165+ if isLeaf {
166+ ans += l
167+ }
168+ return ans
169+ }
170+ ```
72171
172+ ### ** C++**
173+
174+ ``` cpp
175+ struct Trie {
176+ Trie* children[ 26] = { nullptr };
177+ };
178+
179+ class Solution {
180+ public:
181+ int minimumLengthEncoding(vector<string >& words) {
182+ auto root = new Trie();
183+ for (auto& w : words) {
184+ auto cur = root;
185+ for (int i = w.size() - 1; i >= 0; --i) {
186+ if (cur->children[ w[ i] - 'a'] == nullptr) {
187+ cur->children[ w[ i] - 'a'] = new Trie();
188+ }
189+ cur = cur->children[ w[ i] - 'a'] ;
190+ }
191+ }
192+ return dfs(root, 1);
193+ }
194+
195+ private:
196+ int dfs(Trie* cur, int l) {
197+ bool isLeaf = true;
198+ int ans = 0;
199+ for (int i = 0; i < 26; ++i) {
200+ if (cur->children[ i] != nullptr) {
201+ isLeaf = false;
202+ ans += dfs(cur->children[ i] , l + 1);
203+ }
204+ }
205+ if (isLeaf) {
206+ ans += l;
207+ }
208+ return ans;
209+ }
210+ };
73211```
74212
75213### **...**
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