211. Add and Search Word - Data structure design Design a data structure that supports the following two operations:
代码语言:javascript复制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.
Example:
代码语言:javascript复制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
.
思路:
由于这一题的匹配是可以直接使用
.
代替任意字符,所以可以用一个map存下左右的输入字符串,然后用简单的单个字符的比较来做search,但是这里使用前缀树来做。
代码:
go:
代码语言:javascript复制const R = 26
type WordDictionary struct {
Children [R]*WordDictionary
Word bool
}
/** Initialize your data structure here. */
func Constructor() WordDictionary {
return WordDictionary{}
}
/** Adds a word into the data structure. */
func (this *WordDictionary) AddWord(word string) {
cur := this
for i := 0; i < len(word); i {
c := rune(word[i])
if cur.Children[c - 'a'] == nil {
cur.Children[c - 'a'] = &WordDictionary{Children : [R]*WordDictionary{}, Word:false}
}
cur = cur.Children[c - 'a']
}
cur.Word = true
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
func (this *WordDictionary) Search(word string) bool {
return searchWord(word, this)
}
func searchWord(word string, node *WordDictionary) bool {
for i := 0; i < len(word) && node != nil; i {
c := rune(word[i])
if c != '.' {
node = node.Children[c - 'a']
} else {
temp := node
for _, children := range temp.Children {
node = children
if searchWord(word[i 1:], node) {
return true
}
}
}
}
return node != nil && node.Word
}
/**
* Your WordDictionary object will be instantiated and called as such:
* obj := Constructor();
* obj.AddWord(word);
* param_2 := obj.Search(word);
*/