【剑指 Offer II】 063. 替换单词
题目:
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 1:
输入:dictionary = [“cat”,“bat”,“rat”], sentence = “the cattle was rattled by the battery”
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。
注意:本题与主站 648 题相同: https://leetcode-cn.com/problems/replace-words/
答案:
class Solution {
//设计前缀树的数据结构
public class TrieNode{
boolean isEnd;
TrieNode[] next;
TrieNode(){
next = new TrieNode[26];
}
}
public String replaceWords(List<String> dictionary, String sentence) {
//使用前缀树
//将字典插入前缀树
TrieNode root = new TrieNode();
for(int i = 0; i < dictionary.size(); i++){
TrieNode node = root;
for(char c : dictionary.get(i).toCharArray()){
if(node.next[c - 'a'] == null){
node.next[c - 'a'] = new TrieNode();
}
node = node.next[c - 'a'];
}
node.isEnd = true;
}
//判断sentence
String[] strs = sentence.split(" ");
StringBuffer sb = new StringBuffer();
for(String str : strs){
TrieNode node = root;
StringBuffer temp = new StringBuffer();
for(char c : str.toCharArray()){
if(node.next[c - 'a'] == null){
sb.append(str);
temp = new StringBuffer();
break;
}else{
temp.append(c);
//System.out.println(c);
if(node.next[c - 'a'].isEnd == true) break;
node = node.next[c - 'a'];
}
}
sb.append(temp);
sb.append(" ");
}
sb.deleteCharAt(sb.length() - 1);
return sb.toString();
}
}