[Swift]LeetCode1048.最长字符串链 | Longest String Chain

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Given a list of words, each word consists of English lowercase letters.

Let's say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1to make it equal to word2.  For example, "abc" is a predecessor of "abac".

A word chain is a sequence of words [word_1, word_2, ..., word_k] with k >= 1, where word_1 is a predecessor of word_2, word_2 is a predecessor of word_3, and so on.

Return the longest possible length of a word chain with words chosen from the given list of words.

Example 1:

Input: ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: one of the longest word chain is "a","ba","bda","bdca".

Note:

1. 1 <= words.length <= 1000
2. 1 <= words[i].length <= 16
3. words[i] only consists of English lowercase letters.

---

给出一个单词列表，其中每个单词都由小写英文字母组成。

如果我们可以在 word1 的任何地方添加一个字母使其变成 word2，那么我们认为 word1 是 word2 的前身。例如，"abc" 是 "abac" 的前身。

词链是单词 [word_1, word_2, ..., word_k] 组成的序列，k >= 1，其中 word_1 是 word_2 的前身，word_2 是 word_3 的前身，依此类推。

从给定单词列表 words 中选择单词组成词链，返回词链的最长可能长度。

示例：

输入：["a","b","ba","bca","bda","bdca"]
输出：4
解释：最长单词链之一为 "a","ba","bda","bdca"。

提示：

1. 1 <= words.length <= 1000
2. 1 <= words[i].length <= 16
3. words[i] 仅由小写英文字母组成。

---

204ms

class Solution {
    func longestStrChain(_ words: [String]) -> Int {
        var wordsByLength = Array(repeating: [String: Int](), count: 17)
        
        for word in words {
            wordsByLength[word.count][word] = 1
        }
        
        var returnValue = 1
        
        for i in 1...16 {
            let keys = wordsByLength[i].keys
            for word in keys {
                for j in 0..<word.count {
                    var prevKey = word
                    let index = prevKey.index(prevKey.startIndex, offsetBy: j)
                    prevKey.remove(at: index)
                    if let result = wordsByLength[prevKey.count][prevKey] {
                        let this = result + 1
                        wordsByLength[word.count][word] = this
                        returnValue = max(returnValue, this)
                    }
                }
            }
        }
        
        return returnValue
    }
}

---

696ms

class Solution {
    var result = 1
    func longestStrChain(_ words: [String]) -> Int {
        guard words.count > 0 else {
            return 0
        }
        var memo = [String: Int]()
        var dict = [Int: [String]]()
        for word in words {
            dict[word.count] = dict[word.count, default:[String]()]
            dict[word.count]!.append(word)
            memo[word] = 1
        }
        var sortedWords = words.sorted(by: {$0.count < $1.count})
        
        for word in sortedWords {
            for i in 0..<word.count {
                var chars = Array(word)
                chars.remove(at: i)
                let last = String(chars)
                if let value = dict[last.count], value.contains(last) {
                    memo[word] = max(memo[word]!, memo[last]! + 1)
                    result = max(memo[word]!, result)
                }
            }
        }
        return result
    }
}

---

Runtime: 2092 ms

Memory Usage: 21 MB

class Solution {
    func longestStrChain(_ words: [String]) -> Int {
        var words = words
        words.sort(by:{
            return $0.count < $1.count
        })
        let n:Int = words.count
        var dp:[Int] = [Int](repeating:0,count:n)
        dp[0] = 1
        for i in 1..<n
        {
            dp[i] = 1
            for j in 0..<i
            {
                if predecessor(words[j], words[i])
                {
                    dp[i] = max(dp[i], dp[j] + 1)
                }
                
            }
        }
        return dp.max() ?? 0
    }
    
    func predecessor(_ t:String,_ s:String) -> Bool
    {
        let m:Int = t.count
        let n:Int = s.count
        if m != n - 1 {return false}
        var cnt:Int = 0
        var i:Int = 0
        var j:Int = 0
        let arrT:[Character] = Array(t)
        let arrS:[Character] = Array(s)
        while(i < m && j < n)
        {
            if arrT[i] == arrS[j]
            {
                i += 1
                j += 1
            }
            else
            {
                if cnt > 0 {return false}
                cnt += 1
                j += 1
            }
        }
        return true
    }
}