[Leetcode] 648. Replace Words 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79154305

本文介绍了一种使用字典树（Trie）结构实现的算法，该算法能够在给定的句子中找到并替换由一系列根词汇构成的继承词汇，用最短的根词汇进行替代。

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题目：

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 rootforming 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:

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

思路：

又是一道字典树的题目，分为两步：1）将字典中的所有前缀都加入到字典树中。2）对于sentence中的每个单词，我们在字典树中查找是否有它的前缀存在于字典树中，如果有，则用其最小的前缀来代替它；否则就保持不变。最后返回修改后的sentence即可。

代码：

class Solution {
public:
    string replaceWords(vector<string>& dict, string sentence) {
        root = new TrieNode();      // step 1: add all the string in dict
        for (auto &s : dict) {
            addTrieNode(s);
        }
        // step 2: replace the s once it has prefix in the Trie
        istringstream iss(sentence);
        string s, ret, prefix;
        while(iss >> s) {
            prefix = search(s);
            ret += prefix == "" ? s : prefix;
            ret += ' ';
        }
        ret.pop_back();
        return ret;
    }
private:
    struct TrieNode {
        bool finished;
        vector<TrieNode*> children;
        TrieNode() {
            finished = false;
            children = vector<TrieNode*>(26, NULL);
        }
    };
    void addTrieNode(string &s) {
        TrieNode *node = root;
        for (int i = 0; i < s.length(); ++i) {
            int index = s[i] - 'a';
            if (node->children[index] == NULL) {
                node->children[index] = new TrieNode();
            }
            node = node->children[index];
        }
        node->finished = true;
    }
    string search(string &s) {
        TrieNode *node = root;
        string ret;
        for (int i = 0; i < s.length(); ++i) {
            int index = s[i] - 'a';
            if (node->children[index] == NULL) {
                return "";
            }
            ret += ('a' + index);
            node = node->children[index];
            if (node->finished == true) {
                return ret;
            }
        }
        return "";
    }
    TrieNode *root;
};