820. 单词的压缩编码

题目描述

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

给你一个单词数组 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 <= 2000
1 <= words[i].length <= 7
words[i] 仅由小写字母组成

解法

方法一：前缀树

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。

Python3JavaC++Go

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 ans

class 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
}

方法二

Python3JavaC++Go

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
}

820. 单词的压缩编码

题目描述

解法

方法一：前缀树

方法二

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