单词替换

2021-01-29

在英语中，我们有一个叫做 词根(root)的概念，它可以跟着其他一些词组成另一个较长的单词——我们称这个词为 继承词(successor)。例如，词根an，跟随着单词 other(其他)，可以形成新的单词 another(另一个)。

现在，给定一个由许多词根组成的词典和一个句子。你需要将句子中的所有继承词用词根替换掉。如果继承词有许多可以形成它的词根，则用最短的词根替换它。

你需要输出替换之后的句子。

 

示例 1：
输入：dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输出："the cat was rat by the bat"

示例 2：
输入：dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"
输出："a a b c"

示例 3：
输入：dictionary = ["a", "aa", "aaa", "aaaa"], sentence = "a aa a aaaa aaa aaa aaa aaaaaa bbb baba ababa"
输出："a a a a a a a a bbb baba a"

示例 4：
输入：dictionary = ["catt","cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输出："the cat was rat by the bat"

示例 5：
输入：dictionary = ["ac","ab"], sentence = "it is abnormal that this solution is accepted"
输出："it is ab that this solution is ac"

提示：
1 <= dictionary.length <= 1000
1 <= dictionary[i].length <= 100
dictionary[i] 仅由小写字母组成。
1 <= sentence.length <= 10^6
sentence 仅由小写字母和空格组成。
sentence 中单词的总量在范围 [1, 1000] 内。
sentence 中每个单词的长度在范围 [1, 1000] 内。
sentence 中单词之间由一个空格隔开。
sentence 没有前导或尾随空格。

作者：力扣 (LeetCode)
链接：https://leetcode-cn.com/leetbook/read/trie/x0cga2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

字典树（暴力）

java split 空格使用

1	String[] words = sentence.split("\\s+");

class Solution {
    public String replaceWords(List<String> dictionary, String sentence) {

        Trie root = new Trie();
        for (String s : dictionary) {
            root.insert(s);
        }
        String[] words = sentence.split("\\s+");

        StringBuilder bd = new StringBuilder();

        for (int i = 0; i < words.length; i++) {
            boolean flag = false;
            if (root.startWith(words[i].substring(0, 1))) {
                for (int j = 0; j < words[i].length(); j++) {
                    String pWord = words[i].substring(0, j + 1);
                    if (root.search(pWord)) {
                        bd.append(pWord);
                        bd.append(" ");
                        flag = true;
                        break;
                    }
                }
            }
            if (!flag) {
                bd.append(words[i]);
                bd.append(" ");
            }


        }
        if (bd.length() > 0) bd.deleteCharAt(bd.length() - 1);
        return bd.toString();
    }

    private class Trie {
        private boolean isWord = false;
        private Trie[] next = new Trie[26];

        public Trie() {

        }

        public void insert(String word) {
            Trie root = this;
            char[] w = word.toCharArray();
            for (int i = 0; i < w.length; i++) {
                int index = w[i] - 'a';
                if (root.next[index] == null) root.next[index] = new Trie();
                root = root.next[index];
            }
            root.isWord = true;
        }

        public boolean search(String word) {
            Trie root = this;
            char[] w = word.toCharArray();
            for (int i = 0; i < w.length; i++) {
                int index = w[i] - 'a';
                if (root.next[index] == null) return false;
                root = root.next[index];
            }
            return root.isWord;
        }

        public boolean startWith(String prefix) {
            Trie root = this;
            char[] w = prefix.toCharArray();
            for (int i = 0; i < w.length; i++) {
                int index = w[i] - 'a';
                if (root.next[index] == null) return false;
                root = root.next[index];
            }
            return true;
        }

    }
}

哈希表

class Solution {
    public String replaceWords(List<String> roots, String sentence) {
        Set<String> rootset = new HashSet();
        for (String root: roots) rootset.add(root);

        StringBuilder ans = new StringBuilder();
        for (String word: sentence.split("\\s+")) {
            String prefix = "";
            for (int i = 1; i <= word.length(); ++i) {
                prefix = word.substring(0, i);
                if (rootset.contains(prefix)) break;
            }
            if (ans.length() > 0) ans.append(" ");
            ans.append(prefix);
        }
        return ans.toString();
    }
}

作者：LeetCode
链接：https://leetcode-cn.com/problems/replace-words/solution/dan-ci-ti-huan-by-leetcode/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

字典树

class Solution {
    public String replaceWords(List<String> roots, String sentence) {
        TrieNode trie = new TrieNode();
        for (String root: roots) {
            TrieNode cur = trie;
            for (char letter: root.toCharArray()) {
                if (cur.children[letter - 'a'] == null)
                    cur.children[letter - 'a'] = new TrieNode();
                cur = cur.children[letter - 'a'];
            }
            cur.word = root;
        }

        StringBuilder ans = new StringBuilder();

        for (String word: sentence.split("\\s+")) {
            if (ans.length() > 0)
                ans.append(" ");

            TrieNode cur = trie;
            for (char letter: word.toCharArray()) {
                if (cur.children[letter - 'a'] == null || cur.word != null)
                    break;
                cur = cur.children[letter - 'a'];
            }
            ans.append(cur.word != null ? cur.word : word);
        }
        return ans.toString();
    }
}

class TrieNode {
    TrieNode[] children;
    String word;
    TrieNode() {
        children = new TrieNode[26];
    }
}

作者：LeetCode
链接：https://leetcode-cn.com/problems/replace-words/solution/dan-ci-ti-huan-by-leetcode/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。