C++实现LeetCode(648.替换单词)

时间:2022-06-10 07:30:16

[LeetCode] 648.Replace Words 替换单词

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 the root 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"

Note:

  1. The input will only have lower-case letters.
  2. 1 <= dict words number <= 1000
  3. 1 <= sentence words number <= 1000
  4. 1 <= root length <= 100
  5. 1 <= sentence words length <= 1000

这道题给了我们一个前缀字典,又给了一个句子,让我们将句子中较长的单词换成其前缀(如果在前缀字典中存在的话)。我们对于句子中的一个长单词如何找前缀呢,是不是可以根据第一个字母来快速定位呢,比如cattle这个单词的首字母是c,那么我们在前缀字典中找所有开头是c的前缀,为了方便查找,我们将首字母相同的前缀都放到同一个数组中,总共需要26个数组,所以我们可以定义一个二维数组来装这些前缀。还有,我们希望短前缀在长前缀的前面,因为题目中要求用最短的前缀来替换单词,所以我们可以先按单词的长度来给所有的前缀排序,然后再依次加入对应的数组中,这样就可以保证短的前缀在前面。

下面我们就要来遍历句子中的每一个单词了,由于C++中没有split函数,所以我们就采用字符串流来提取每一个单词,对于遍历到的单词,我们根据其首字母查找对应数组中所有以该首字母开始的前缀,然后直接用substr函数来提取单词中和前缀长度相同的子字符串来跟前缀比较,如果二者相等,说明可以用前缀来替换单词,然后break掉for循环。别忘了单词之前还要加上空格,参见代码如下:

解法一:

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class Solution {
public:
    string replaceWords(vector<string>& dict, string sentence) {
        string res = "", t = "";
        vector<vector<string>> v(26);
        istringstream is(sentence);
        sort(dict.begin(), dict.end(), [](string &a, string &b) {return a.size() < b.size();});
        for (string word : dict) {
            v[word[0] - 'a'].push_back(word);
        }
        while (is >> t) {
            for (string word : v[t[0] - 'a']) {
                if (t.substr(0, word.size()) == word) {
                    t = word;
                    break;
                }
            }
            res += t + " ";
        }
        res.pop_back();
        return res;
    }
};

你以为想出了上面的解法,这道题就算做完了?? Naive! ! ! 这道题最好的解法其实是用前缀树(Trie / Prefix Tree)来做,关于前缀树使用之前有一道很好的入门题Implement Trie (Prefix Tree)。了解了前缀树的原理机制,那么我们就可以发现这道题其实很适合前缀树的特点。我们要做的就是把所有的前缀都放到前缀树里面,而且在前缀的最后一个结点的地方将标示isWord设为true,表示从根节点到当前结点是一个前缀,然后我们在遍历单词中的每一个字母,我们都在前缀树查找,如果当前字母对应的结点的表示isWord是true,我们就返回这个前缀,如果当前字母对应的结点在前缀树中不存在,我们就返回原单词,这样就能完美的解决问题了。所以啊,以后遇到了有关前缀或者类似的问题,一定不要忘了前缀树这个神器哟~

解法二:

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class Solution {
public:
    class TrieNode {
    public:
        bool isWord;
        TrieNode *child[26];
        TrieNode(): isWord(false) {
            for (auto &a : child) a = NULL;
        }
    };
    
    string replaceWords(vector<string>& dict, string sentence) {
        string res = "", t = "";
        istringstream is(sentence);
        TrieNode *root = new TrieNode();
        for (string word : dict) {
            insert(root, word);
        }
        while (is >> t) {
            if (!res.empty()) res += " ";
            res += findPrefix(root, t);
        }
        return res;
    }
    
    void insert(TrieNode* node, string word) {
        for (char c : word) {
            if (!node->child[c - 'a']) node->child[c - 'a'] = new TrieNode();
            node = node->child[c - 'a'];
        }
        node->isWord = true;
    }
    
    string findPrefix(TrieNode* node, string word) {
        string cur = "";
        for (char c : word) {
            if (!node->child[c - 'a']) break;
            cur.push_back(c);
            node = node->child[c - 'a'];
            if (node->isWord) return cur;
        }
        return word;
    }
};

类似题目:

Implement Trie (Prefix Tree)

参考资料:

https://discuss.leetcode.com/topic/97203/trie-tree-concise-java-solution-easy-to-understand

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原文链接:https://www.cnblogs.com/grandyang/p/7423420.html