设计一个支持以下两种操作的数据结构:
void addWord(word)
bool search(word)
search(word) 可以搜索文字或正则表达式字符串,字符串只包含字母 . 或 a-z 。 . 可以表示任何一个字母。
示例:
addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true
说明:
你可以假设所有单词都是由小写字母 a-z 组成的。
思路就是,字典树,加DFS递归,一定注意递归退出条件,我因为退出条件的部分没想明白,
把自己绕进死胡同,耗费大量时间了。
#include <iostream>
#include <string.h>
#include <vector>
#include <set>
using namespace std;
class WordDictionary {
private:
int MaxWordLen;
int MinWordLen;
int CurWordLen;
WordDictionary* next[26];
bool isEnd;
public:
/** Initialize your data structure here. */
WordDictionary() {
MaxWordLen = 0;
MinWordLen = INT_MAX;
CurWordLen = 0;
memset(next,0x00,sizeof(next));
isEnd = false;
}
/** Adds a word into the data structure. */
void addWord(string word) {
if(word.length()==0)
return;
if(word.length()>MaxWordLen)
MaxWordLen = word.length();
if(word.length()<MinWordLen)
MinWordLen = word.length();
WordDictionary *node = this;
for(char c:word){
if(node->next[c-'a']==NULL){
node->next[c-'a'] = new WordDictionary();
}
node = node->next[c-'a'];
}
node->isEnd = true;
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
bool search(string word) {
if(word.length()>MaxWordLen || word.length()<MinWordLen){
return false;
}
CurWordLen = word.length();
const char *charArry =word.c_str();
//printf("[%s]\n",charArry);
WordDictionary *node = this;
return IsLegalTo(node,charArry,0);
}
bool IsLegalTo(WordDictionary *node,const char*arr, int index){
// cout<<"arr:"<<arr[index]<<"----index:"<<index<<"---CurWordLen:"<<CurWordLen<<endl;
if(node==NULL){
return false;
}
if(index==CurWordLen){
//cout<<"node->isEnd:"<<node->isEnd<<endl;
return node->isEnd;
}
if(arr[index]=='.'){
for(int i=0;i<26;i++){
if(IsLegalTo(node->next[i],arr,index+1)){
return true;
}
}
//cout<<"get in this"<<endl;
return false;
}else{
char c = arr[index];
if(node->next[c-'a']==NULL){
//cout<<"char:"<<arr[index]<<"--index:"<<index<<endl;
return false;
}
else{
node =node->next[c-'a'];
return IsLegalTo(node,arr,index+1);
}
}
return false;
}
};
int main(){
WordDictionary *pw = new WordDictionary();
//test case 1
// pw->addWord("a");
// pw->addWord("ab");
// cout<<"main:"<<pw->search("a")<<endl;
// cout<<"main:"<<pw->search("a.")<<endl;
// cout<<"main:"<<pw->search("ab")<<endl;
// cout<<"main:"<<pw->search(".a")<<endl;
// cout<<"main:"<<pw->search(".b")<<endl;
// cout<<"main:"<<pw->search("ab.")<<endl;
// cout<<"main:"<<pw->search(".")<<endl;
// cout<<"main:"<<pw->search("..")<<endl;
//test case 2
pw->addWord("bad");
pw->addWord("dad");
pw->addWord("mad");
cout<<pw->search("pad")<<endl;
cout<<pw->search("bad")<<endl;
cout<<pw->search(".ad")<<endl;
cout<<pw->search("b..")<<endl;
// test case 3
// pw->addWord("at");
// pw->addWord("and");
// pw->addWord("an");
// pw->addWord("add");
// cout<<"main:"<<pw->search("a")<<endl;
// cout<<"main:"<<pw->search(".at")<<endl;
// pw->addWord("bat");
// cout<<"addword(bat)++++++++++++++++++++"<<endl;
// cout<<"main:"<<pw->search(".at")<<endl;
// cout<<"main:"<<pw->search("an.")<<endl;
// cout<<"main:"<<pw->search("a.d.")<<endl;
// cout<<"main:"<<pw->search("b.")<<endl;
// cout<<"main:"<<pw->search("a.d")<<endl;
// cout<<"main:"<<pw->search(".")<<endl;
// test case 4
// pw->addWord("ran");
// pw->addWord("rune");
// pw->addWord("runner");
// pw->addWord("runs");
// pw->addWord("add");
// pw->addWord("adds");
// pw->addWord("adder");
// pw->addWord("addee");
// cout<<"main:"<<pw->search("....e.")<<endl;
// test case 4
// pw->addWord("ran");
// pw->addWord("rune");
// pw->addWord("runner");
// pw->addWord("runs");
// pw->addWord("add");
// pw->addWord("adds");
// pw->addWord("adder");
// pw->addWord("addee");
// cout<<"main:"<<pw->search(".an.")<<endl;
return 0;
}