leetcode-648-Replace-Words

发表于   |  

描述

In English, we have a concept called root, which can be followed by some other words to form another longer word - let’s call this word successor. For example, the root an, followed by other, which can form another word another.

Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor in the sentence with the root forming it. If a successor has many roots can form it, replace it with the root with the shortest length.

You need to output the sentence after the replacement.

Example 1:

1
2
3
Input: dict = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"
Output: "the cat was rat by the bat"

Note:

  1. The input will only have lower-case letters.
  2. 1 <= dict words number <= 1000
  3. 1 <= sentence words number <= 1000
  4. 1 <= root length <= 100
  5. 1 <= sentence words length <= 1000

分析

Trie 的应用，根据前缀替换单词。时间复杂度 $O(n)$，空间复杂度 $O(n)$。

解决方案1(Java)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
class Solution {
    
    public String replaceWords(List<String> dict, String sentence) {
        TrieNode trie = new TrieNode();
        for (String word: dict) {
            TrieNode current = trie;
            for (char c: word.toCharArray()) {
                if (!current.children.containsKey(c)) {
                    current.children.put(c, new TrieNode());
                }
                current = current.children.get(c);
            }
            current.word = word;
        }
        
        StringBuilder result = new StringBuilder();
        for (String word: sentence.split("\\s+")) {
            if (result.length() > 0) {
                result.append(" ");
            }
            
            TrieNode current = trie;
            
            for (char c: word.toCharArray()) {
                if (!current.children.containsKey(c) || current.word != null) {
                    break;
                }
                current = current.children.get(c);
            }
            result.append(current.word != null ? current.word: word);
        }
        return result.toString();
    }
}

class TrieNode {
    String word;
    Map<Character, TrieNode> children = new HashMap();
}

相关问题

(M) Implement Trie (Prefix Tree)

题目来源