LeetCode 648. Replace Words 字典樹練習

In English, we have a concept called root, which can be followed by some other words to form another longer word - let's call this word successor. For example, the root an, followed by other, which can form another word another.

Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor in the sentence with theroot forming it. If a successor has many roots can form it, replace it with the root with the shortest length.

You need to output the sentence after the replacement.

Example 1:

Input: dict = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"
Output: "the cat was rat by the bat"

題意：用字典中存在的前綴代替句子中的單詞，若有多個前綴可以表示單詞，則選擇最短的一個

用基礎字典樹即可解決，將字典中的前綴字符串存入字典樹中，遍歷每個句子的單詞，找到最短的前綴返回即可

字典樹結構：

struct Trie{
	int val;	//表示此位置是否有前綴結束
	vector<Trie*> child;
	Trie():child(vector<Trie*>(26,NULL)),val(-1){}
};

首先是字典樹的創建（插入）過程，題目只要返回最短的前綴字符串，在插入時對於每一個字符串的最後一個字符節點的val設置爲0表示此位置爲結束位置。

void insert(string s,Trie* root){
	for(int i=0;i<s.size();i++){
		int num = s[i]-'a';
		if(root->child[num] == NULL)
		{
			root->child[num] = new Trie();
		}
		root = root->child[num];
        if(i!=s.size()-1&&root->val!=0)    	
            root->val = 1;
        else root->val = 0;	//若此位置爲結束位置，將val設置爲0
	}
}
   

查找時，遇到val爲0的字符節點就可以返回，這樣就能保證返回的前綴爲最短的，若字典樹中不存在此單詞的前綴，那麼查找停止位置的val不等於0；

    string search(string s,Trie* root){
	string res = "";
	for(int i=0;i<s.size();i++){
		int num = s[i] -'a';
		if(root->child[num] == NULL)
			break;
        root = root->child[num];
		    res += s[i];
        if(root->val == 0)
            return res;
		}
        if(root->val == 0)
	        return res;
        else return s;
}

在分割句子時可以使用強大的stringstream，將sentence先讀入字符串流中，以空格爲標示分割句子，超級方便，字符串分割利器，還能類型轉化等等：

string replaceWords(vector<string>& dict, string sentence) {
        Trie *root = new Trie();
        string res = "";
        for(string s:dict)
            insert(s,root);
        stringstream ss(sentence);
        string s;
        int sign = 1;
        while(ss>>s){
            string temp = search(s,root);
            if(sign!=1)
                res += " ";
            if(temp.empty())
                res +=s;
            else res +=temp;
            sign++;
        }
        return res;
    }