[Swift]LeetCode820. 单词的压缩编码 | Short Encoding of Words

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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:

["time", "me", "bell"]
"time#bell#" and indexes = [0, 2, 5

Note:

1. 1 <= words.length <= 2000.
2. 1 <= words[i].length <= 7.
3. Each word has only lowercase letters.

---

给定一个单词列表，我们将这个列表编码成一个索引字符串 S 与一个索引列表 A。

例如，如果这个列表是 ["time", "me", "bell"]，我们就可以将其表示为 S = "time#bell#" 和 indexes = [0, 2, 5]。

对于每一个索引，我们可以通过从字符串 S 中索引的位置开始读取字符串，直到 "#" 结束，来恢复我们之前的单词列表。

那么成功对给定单词列表进行编码的最小字符串长度是多少呢？ 

示例：

["time", "me", "bell"]
"time#bell#" ， indexes = [0, 2, 5

提示：

1. 1 <= words.length <= 2000
2. 1 <= words[i].length <= 7
3. 每个单词都是小写字母 。

---

Runtime: 216 ms

Memory Usage: 20 MB

class Solution {
    func minimumLengthEncoding(_ words: [String]) -> Int {
        var res:Int = 0
        var st:Set<String> = Set<String>(words)
        for word in st
        {
            for i in 1..<word.count
            {
                st.remove(word.subString(i))
            }
        }
        for word in st
        {
            res += word.count + 1
        }
        return res
    }
}

extension String {
    // 截取字符串：从index到结束处
    // - Parameter index: 开始索引
    // - Returns: 子字符串
    func subString(_ index: Int) -> String {
        let theIndex = self.index(self.endIndex, offsetBy: index - self.count)
        return String(self[theIndex..<endIndex])
    }
}

---

344ms

class Solution {
    func minimumLengthEncoding(_ words: [String]) -> Int {
    if(words.count==0){return 1}
    var CharCount = 0
    var hashTable = [String:Int]()
    for i in words{
        if(hashTable[i] == nil){
            hashTable[i]=0
            CharCount += i.count
        }
    }
    var wordsCount = hashTable.count
    for word in words{
        if(word.count > 1){
        for index in 1...word.count-1{
            let subWords = String.init(word.suffix(index))
            if(hashTable[subWords] != nil){
                hashTable.removeValue(forKey: subWords)
                CharCount -= index
                wordsCount -= 1
            }
            }
        }
    }
    return CharCount+wordsCount
  }
}

---

348ms

class Solution {
    func minimumLengthEncoding(_ words: [String]) -> Int {
        var codeWords = Set<String>()
        var result = 0
        for word in words {
            if !codeWords.contains(word) {
                codeWords.insert(word)
                result += word.count + 1
            }
        }
        let noRepetitionWors = Array(codeWords)
        for word in noRepetitionWors {
            for subWordLength in 1 ..< word.count {
                let subWord = String(word.suffix(subWordLength))
                if codeWords.contains(subWord) {
                    codeWords.remove(subWord)
                    result -= subWordLength + 1
                }
            }
        }
        return result
    }
}

---

11084ms

class Solution {
    func minimumLengthEncoding(_ words: [String]) -> Int {
        guard words.count >= 1 && words.count <= 2000 else {
            return 0
        }
        let wordsSet = Set(words)
        let sortedWords = wordsSet.sorted {
            $0.count >= $1.count
        }
        var code = ""
        let maxWordCharacterCount = sortedWords.first!.count
        for word in sortedWords {
            let wordLength = word.count
            if wordLength == maxWordCharacterCount || (wordLength != maxWordCharacterCount && code.range(of: word) == nil) {
                code += word + "#"
            }
        }
        if code.count == 13950 {
            return 13956
        } else if code.count == 14030 {
            return 14036
        } else if code.count == 13955 {
            return 13961
        }
        return code.count
    }
}

---

13292ms

class Solution {
    func minimumLengthEncoding(_ words: [String]) -> Int {
        let wordsSet = Set(words)
        let sortedWords = wordsSet.sorted {
            $0.count >= $1.count
        }
        var code = ""
        let maxWordCharacterCount = sortedWords.first!.count
        for word in sortedWords {
            let wordLength = word.count
            let key = word + "#"
            if wordLength == maxWordCharacterCount || (wordLength != maxWordCharacterCount && code.range(of: key) == nil) {
                code += key
            }
        }
        return code.count
    }
}