单词数组 words 的 有效编码 由任意助记字符串 s 和下标数组 indices 组成，且满足：
words.length == indices.length- 助记字符串
s以'#'字符结尾 - 对于每个下标
indices[i]，s的一个从indices[i]开始、到下一个'#'字符结束（但不包括'#'）的 子字符串 恰好与words[i]相等
给定一个单词数组 words ，返回成功对 words 进行编码的最小助记字符串 s 的长度 。
示例 1：
输入：words = ["time", "me", "bell"]
输出：10
解释：一组有效编码为 s = "time#bell#" 和 indices = [0, 2, 5] 。
words[0] = "time" ，s 开始于 indices[0] = 0 到下一个 '#' 结束的子字符串，如加粗部分所示 "time#bell#"
words[1] = "me" ，s 开始于 indices[1] = 2 到下一个 '#' 结束的子字符串，如加粗部分所示 "time#bell#"
words[2] = "bell" ，s 开始于 indices[2] = 5 到下一个 '#' 结束的子字符串，如加粗部分所示 "time#bell#"
示例 2：
输入：words = ["t"] 输出：2 解释：一组有效编码为 s = "t#" 和 indices = [0] 。
提示：
1 <= words.length <= 20001 <= words[i].length <= 7words[i]仅由小写字母组成
注意：本题与主站 820 题相同： https://leetcode.cn/problems/short-encoding-of-words/
题目大意：充分利用重叠的后缀，使有效编码尽可能短。
判断当前单词是否是其他单词的后缀，若是，就不用写入助记字符串中，否则需要写入并且加上一个 # 后缀。
class Trie:
def __init__(self) -> None:
self.children = [None] * 26
class Solution:
def minimumLengthEncoding(self, words: List[str]) -> int:
root = Trie()
for w in words:
cur = root
for c in w[::-1]:
idx = ord(c) - ord("a")
if cur.children[idx] == None:
cur.children[idx] = Trie()
cur = cur.children[idx]
return self.dfs(root, 1)
def dfs(self, cur: Trie, l: int) -> int:
isLeaf, ans = True, 0
for i in range(26):
if cur.children[i] != None:
isLeaf = False
ans += self.dfs(cur.children[i], l + 1)
if isLeaf:
ans += l
return ansclass Trie {
Trie[] children = new Trie[26];
}
class Solution {
public int minimumLengthEncoding(String[] words) {
Trie root = new Trie();
for (String w : words) {
Trie cur = root;
for (int i = w.length() - 1; i >= 0; i--) {
int idx = w.charAt(i) - 'a';
if (cur.children[idx] == null) {
cur.children[idx] = new Trie();
}
cur = cur.children[idx];
}
}
return dfs(root, 1);
}
private int dfs(Trie cur, int l) {
boolean isLeaf = true;
int ans = 0;
for (int i = 0; i < 26; i++) {
if (cur.children[i] != null) {
isLeaf = false;
ans += dfs(cur.children[i], l + 1);
}
}
if (isLeaf) {
ans += l;
}
return ans;
}
}struct Trie {
Trie* children[26] = {nullptr};
};
class Solution {
public:
int minimumLengthEncoding(vector<string>& words) {
auto root = new Trie();
for (auto& w : words) {
auto cur = root;
for (int i = w.size() - 1; i >= 0; --i) {
if (cur->children[w[i] - 'a'] == nullptr) {
cur->children[w[i] - 'a'] = new Trie();
}
cur = cur->children[w[i] - 'a'];
}
}
return dfs(root, 1);
}
private:
int dfs(Trie* cur, int l) {
bool isLeaf = true;
int ans = 0;
for (int i = 0; i < 26; ++i) {
if (cur->children[i] != nullptr) {
isLeaf = false;
ans += dfs(cur->children[i], l + 1);
}
}
if (isLeaf) {
ans += l;
}
return ans;
}
};type trie struct {
children [26]*trie
}
func minimumLengthEncoding(words []string) int {
root := new(trie)
for _, w := range words {
cur := root
for i := len(w) - 1; i >= 0; i-- {
if cur.children[w[i]-'a'] == nil {
cur.children[w[i]-'a'] = new(trie)
}
cur = cur.children[w[i]-'a']
}
}
return dfs(root, 1)
}
func dfs(cur *trie, l int) int {
isLeaf, ans := true, 0
for i := 0; i < 26; i++ {
if cur.children[i] != nil {
isLeaf = false
ans += dfs(cur.children[i], l+1)
}
}
if isLeaf {
ans += l
}
return ans
}class Trie:
def __init__(self):
self.children = [None] * 26
def insert(self, w):
node = self
pref = True
for c in w:
idx = ord(c) - ord("a")
if node.children[idx] is None:
node.children[idx] = Trie()
pref = False
node = node.children[idx]
return 0 if pref else len(w) + 1
class Solution:
def minimumLengthEncoding(self, words: List[str]) -> int:
words.sort(key=lambda x: -len(x))
trie = Trie()
return sum(trie.insert(w[::-1]) for w in words)class Trie {
Trie[] children = new Trie[26];
int insert(String w) {
Trie node = this;
boolean pref = true;
for (int i = w.length() - 1; i >= 0; --i) {
int idx = w.charAt(i) - 'a';
if (node.children[idx] == null) {
pref = false;
node.children[idx] = new Trie();
}
node = node.children[idx];
}
return pref ? 0 : w.length() + 1;
}
}
class Solution {
public int minimumLengthEncoding(String[] words) {
Arrays.sort(words, (a, b) -> b.length() - a.length());
int ans = 0;
Trie trie = new Trie();
for (String w : words) {
ans += trie.insert(w);
}
return ans;
}
}class Trie {
public:
vector<Trie*> children;
Trie()
: children(26) {}
int insert(string w) {
Trie* node = this;
bool pref = true;
for (char c : w) {
c -= 'a';
if (!node->children[c]) {
pref = false;
node->children[c] = new Trie();
}
node = node->children[c];
}
return pref ? 0 : w.size() + 1;
}
};
class Solution {
public:
int minimumLengthEncoding(vector<string>& words) {
sort(words.begin(), words.end(), [](string& a, string& b) { return a.size() > b.size(); });
Trie* trie = new Trie();
int ans = 0;
for (auto& w : words) {
reverse(w.begin(), w.end());
ans += trie->insert(w);
}
return ans;
}
};type Trie struct {
children [26]*Trie
}
func newTrie() *Trie {
return &Trie{}
}
func (this *Trie) insert(w string) int {
node := this
pref := true
for i := len(w) - 1; i >= 0; i-- {
idx := w[i] - 'a'
if node.children[idx] == nil {
pref = false
node.children[idx] = newTrie()
}
node = node.children[idx]
}
if pref {
return 0
}
return len(w) + 1
}
func minimumLengthEncoding(words []string) int {
sort.Slice(words, func(i, j int) bool { return len(words[i]) > len(words[j]) })
trie := newTrie()
ans := 0
for _, w := range words {
ans += trie.insert(w)
}
return ans
}