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Prefix and Suffix Search (H)

题目

Design a special dictionary which has some words and allows you to search the words in it by a prefix and a suffix.

Implement the WordFilter class:

• WordFilter(string[] words) Initializes the object with the words in the dictionary.
• f(string prefix, string suffix) Returns the index of the word in the dictionary which has the prefix prefix and the suffix suffix. If there is more than one valid index, return the largest of them. If there is no such word in the dictionary, return -1.

Example 1:

Input
["WordFilter", "f"]
[[["apple"]], ["a", "e"]]
Output
[null, 0]

Explanation
WordFilter wordFilter = new WordFilter(["apple"]);
wordFilter.f("a", "e"); // return 0, because the word at index 0 has prefix = "a" and suffix = ‘e".

Constraints:

• 1 <= words.length <= 15000
• 1 <= words[i].length <= 10
• 1 <= prefix.length, suffix.length <= 10
• words[i], prefix and suffix consist of lower-case English letters only.
• At most 15000 calls will be made to the function f.

题意

给定一个字符串数组，找到下标最大的字符串，使其前缀和后缀与给定值相同。

思路

字典树Trie应用。生成前缀树和后缀树，每次查询时，通过prefix找到所有满足条件的字符串，将其对应的下标加入到优先队列中，通过suffix进行相同操作，最后比较两个优先队列得到答案。

代码实现

Java

class WordFilter {
    private Trie prefixT = new Trie(), suffixT = new Trie();
    private Queue<Integer> prefixQ = new PriorityQueue<>(Comparator.reverseOrder());
    private Queue<Integer> suffixQ = new PriorityQueue<>(Comparator.reverseOrder());

    public WordFilter(String[] words) {
        for (int i = 0; i < words.length; i++) {
            insert(prefixT, words[i].toCharArray(), i);
            insert(suffixT, new StringBuilder(words[i]).reverse().toString().toCharArray(), i);
        }
    }

    public int f(String prefix, String suffix) {
        prefixQ.clear();
        suffixQ.clear();
        search(prefixT, prefixQ, prefix);
        search(suffixT, suffixQ, new StringBuilder(suffix).reverse().toString());

        while (!prefixQ.isEmpty() && !suffixQ.isEmpty()) {
            int a = prefixQ.peek(), b = suffixQ.peek();
            if (a > b) prefixQ.poll();
            else if (a < b) suffixQ.poll();
            else return a;
        }

        return -1;
    }

    private void insert(Trie root, char[] word, int index) {
        Trie cur = root;
        for (char c : word) {
            if (cur.next[c - ‘a‘] == null) cur.next[c - ‘a‘] = new Trie();
            cur = cur.next[c - ‘a‘];
        }
        cur.end = true;
        cur.index = index;
    }

    private void search(Trie root, Queue<Integer> pq, String part) {
        for (char c : part.toCharArray()) {
            if (root.next[c - ‘a‘] == null) return;
            root = root.next[c - ‘a‘];
        }

        Queue<Trie> q = new LinkedList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            Trie cur = q.poll();
            if (cur.end) pq.offer(cur.index);
            for (int i = 0; i < 26; i++) {
                if (cur.next[i] != null) q.offer(cur.next[i]);
            }
        }
    }

    class Trie {
        Trie[] next = new Trie[26];
        boolean end;
        int index;
    }
}

0745. Prefix and Suffix Search (H)

标签：most   integer   case   ever   empty   one   private   init   val   

原文地址：https://www.cnblogs.com/mapoos/p/14724309.html