LeetCode——Add and Search Word - Data structure design

Description:

Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)

search(word) can search a literal word or a regular expression string containing only letters

a-z

or

.

. A

.

means it can represent any one letter.

For example:

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

Note:

You may assume that all words are consist of lowercase letters

a-z

.

题目很好理解：添加单词查找单词。

首先想到的是用线性数据结构，然后逐个匹配查找。可以想象一定是超时的。

public class WordDictionary {

private List<String> arr = new ArrayList<String>();

// Adds a word into the data structure.
public void addWord(String word) {
arr.add(word);
}

// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
public boolean search(String word) {
boolean flag = true;
for(String str : arr) {
if(str.equals(word)) {
flag = true;
break;
}
else if(arr.contains(".")) {
if(word.length() != str.length()) {
flag = false;
break;
}

for(int i=0; i<str.length();) {
char ch1 = str.charAt(i);
char ch2 = word.charAt(i);
if(ch1 == '.' || ch2 == '.' || ch1 == ch2) {
i ++;
}
else {
flag = false;
break;
}
}

}
else {
flag = false;
break;
}
}
return flag;
}
}

// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");

要优化明显要使用Trie树来减少无用的比较次数，从而降低时间复杂度。

关于Trie树：/article/5225757.html

public class WordDictionary {

private TrieNode root;

public WordDictionary() {
this.root = new TrieNode();
}

private class TrieNode {
private TrieNode[] son;
private char val;
private boolean isEnd;
public TrieNode() {
this.son = new TrieNode[26];
this.isEnd = false;
}
}

// Adds a word into the data structure.
public void addWord(String word) {

char[] wordChars = word.toCharArray();
TrieNode node = this.root;

for(char ch : wordChars) {
int pos = ch - 'a';
if(node.son[pos] == null) {
node.son[pos] = new TrieNode();
node.son[pos].val = ch;
}
node = node.son[pos];
}
node.isEnd = true;
}

public boolean patternSearch(String word, TrieNode node) {
char[] wordChars = word.toCharArray();
for(int at=0; at<word.length(); at++) {
char ch = wordChars[at];
if(ch != '.') {
int pos = ch - 'a';
if(node.son[pos] != null) {

node = node.son[pos];
}
else {
return false;
}
}
else {
int flag = 0;
for(int i=0; i<26; i++) {
if(node.son[i] != null) {
boolean b =patternSearch(word.substring(at+1), node.son[i]);
if(b) return b;
else {
flag ++;
}
}
else {
flag ++;
continue;
}
}
if(flag == 26) {
return false;
}
}
}
return node.isEnd;
}

// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
public boolean search(String word) {

return patternSearch(word, this.root);
}

}