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剑指 Offer II 065. 最短的单词编码

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 <= 2000
• 1 <= words[i].length <= 7
• words[i] 仅由小写字母组成

 

注意：本题与主站 820 题相同： https://leetcode-cn.com/problems/short-encoding-of-words/

解法

题目大意：充分利用重叠的后缀，使有效编码尽可能短

Python3

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 i in range(len(w) - 1, -1, -1):
                idx = ord(w[i]) - 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

Java

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;
    }
}

Go

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
}

C++

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;
    }
};

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