0820. Short Encoding of Words

# 820. Short Encoding of Words#

## 题目 #

valid encoding of an array of `words` is any reference string `s` and array of indices `indices` such that:

• `words.length == indices.length`
• The reference string `s` ends with the `'#'` character.
• For each index `indices[i]`, the substring of `s` starting from `indices[i]` and up to (but not including) the next `'#'` character is equal to `words[i]`.

Given an array of `words`, return the length of the shortest reference string `s` possible of any valid encoding of `words`*.*

Example 1:

``````Input: words = ["time", "me", "bell"]
Output: 10
Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"
``````

Example 2:

``````Input: words = ["t"]
Output: 2
Explanation: A valid encoding would be s = "t#" and indices = [0].
``````

Constraints:

• `1 <= words.length <= 2000`
• `1 <= words[i].length <= 7`
• `words[i]` consists of only lowercase letters.

## 题目大意 #

• 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

## 解题思路 #

• 暴力解法。先将所有的单词放入字典中。然后针对字典中的每个单词，逐一从字典中删掉自己的子字符串，这样有相同后缀的字符串被删除了，字典中剩下的都是没有共同前缀的。最终的答案是剩下所有单词用 # 号连接之后的总长度。
• Trie 解法。构建 Trie 树，相同的后缀会被放到从根到叶子节点中的某个路径中。最后依次遍历一遍所有单词，如果单词最后一个字母是叶子节点，说明这个单词是要选择的，因为它可能是包含了一些单词后缀的最长单词。累加这个单词的长度并再加 1(# 字符的长度)。最终累加出来的长度即为题目所求的答案。

## 代码 #

``````package leetcode

// 解法一 暴力
func minimumLengthEncoding(words []string) int {
res, m := 0, map[string]bool{}
for _, w := range words {
m[w] = true
}
for w := range m {
for i := 1; i < len(w); i++ {
delete(m, w[i:])
}
}
for w := range m {
res += len(w) + 1
}
return res
}

// 解法二 Trie
type node struct {
value byte
sub   []*node
}

func (t *node) has(b byte) (*node, bool) {
if t == nil {
return nil, false
}
for i := range t.sub {
if t.sub[i] != nil && t.sub[i].value == b {
return t.sub[i], true
}
}
return nil, false
}

func (t *node) isLeaf() bool {
if t == nil {
return false
}
return len(t.sub) == 0
}

func (t *node) add(s []byte) {
now := t
for i := len(s) - 1; i > -1; i-- {
if v, ok := now.has(s[i]); ok {
now = v
continue
}
temp := new(node)
temp.value = s[i]
now.sub = append(now.sub, temp)
now = temp
}
}

func (t *node) endNodeOf(s []byte) *node {
now := t
for i := len(s) - 1; i > -1; i-- {
if v, ok := now.has(s[i]); ok {
now = v
continue
}
return nil
}
return now
}

func minimumLengthEncoding1(words []string) int {
res, tree, m := 0, new(node), make(map[string]bool)
for i := range words {
if !m[words[i]] {
m[words[i]] = true
}
}
for s := range m {
if tree.endNodeOf([]byte(s)).isLeaf() {
res += len(s)
res++
}
}
return res
}
``````

Apr 8, 2023