描述

在英语中，有一个叫做 词根(root) 的概念，它可以跟着其他一些词组成另一个较长的单词——我们称这个词为 继承词(successor)。例如，词根an，跟随着单词 other(其他)，可以形成新的单词 another(另一个)。

现在，给定一个由许多词根组成的词典和一个句子，需要将句子中的所有继承词用词根替换掉。如果继承词有许多可以形成它的词根，则用最短的词根替换它。

需要输出替换之后的句子。

示例

示例 1：

  1. 输入：dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"
  2. 输出："the cat was rat by the bat"

示例 2：

输入：dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"
输出："a a b c"

示例 3：

输入：dictionary = ["a", "aa", "aaa", "aaaa"], sentence = "a aa a aaaa aaa aaa aaa aaaaaa bbb baba ababa"
输出："a a a a a a a a bbb baba a"

示例 4：

输入：dictionary = ["catt","cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输出："the cat was rat by the bat"

示例 5：

输入：dictionary = ["ac","ab"], sentence = "it is abnormal that this solution is accepted"
输出："it is ab that this solution is ac"

提示

  • 1 <= dictionary.length <= 1000
  • 1 <= dictionary[i].length <= 100
  • dictionary[i] 仅由小写字母组成。
  • 1 <= sentence.length <= 10^6
  • sentence 仅由小写字母和空格组成。
  • sentence 中单词的总量在范围 [1, 1000] 内。
  • sentence 中每个单词的长度在范围 [1, 1000] 内。
  • sentence 中单词之间由一个空格隔开。
  • sentence 没有前导或尾随空格。

解题思路

前缀树

  • 把所有的词根放入前缀树中，在树上查找每个单词的最短词根，该操作可在线性时间内完成。

代码

class Solution {
    public String replaceWords(List<String> dictionary, String sentence) {
        TrieNode trie = new TrieNode();
        for (String dict : dictionary) {
            TrieNode cur = trie;
            for (char letter : dict.toCharArray()) {
                if (cur.children[letter - 'a'] == null) {
                    cur.children[letter - 'a'] = new TrieNode();
                }
                cur = cur.children[letter - 'a'];
            }
            cur.word = dict;
        }
        StringBuilder ans = new StringBuilder();
        for (String word : sentence.split("\\s+")) {
            if (ans.length() > 0) {
                ans.append(" ");
            }
            TrieNode cur = trie;
            for (char letter : word.toCharArray()) {
                if (cur.children[letter - 'a'] == null || cur.word != null) {
                    break;
                }
                cur = cur.children[letter - 'a'];
            }
            ans.append(cur.word != null ? cur.word : word);
        }
        return ans.toString();
    }
}
class TrieNode {
    TrieNode[] children;
    String word;
    TrieNode () {
        children = new TrieNode[26];
    }
}

复杂度分析

时间复杂度：O(N)，其中 N 是 sentence 的长度。每次查询操作为线性时间复杂度。
空间复杂度：O(N)，前缀树的大小。