又一個把性能發揮到極致到例子

Word Search II

給定2D木板和詞典中的單詞列表，請在木板中查找所有單詞。每個單詞必須由順序相鄰的單元格的字母構成，其中“相鄰”單元格是水平或垂直相鄰的單元格。同一字母單元在一個單詞中最多隻能使用一次。

Example:

Input: 
board = [
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]
words = ["oath","pea","eat","rain"]

Output: ["eat","oath"]

Word Search II

Backtracking + Trie

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Intuitively, start from every cell and try to build a word in the dictionary. Backtracking (dfs) is the powerful way to exhaust every possible ways. Apparently, we need to do pruning when current character is not in any word.

1. How do we instantly know the current character is invalid? HashMap?
2. How do we instantly know what’s the next valid character? LinkedList?
3. But the next character can be chosen from a list of characters. "Mutil-LinkedList"?

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Combing them, Trie is the natural choice. Notice that:

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1. TrieNode is all we need. search and startsWith are useless.
2. No need to store character at TrieNode. c.next[i] != null is enough.
3. Never use c1 + c2 + c3. Use StringBuilder.
4. No need to use O(n^2) extra space visited[m][n].
5. No need to use StringBuilder. Storing word itself at leaf node is enough.
6. No need to use HashSet to de-duplicate. Use “one time search” trie.

For more explanations, check out dietpepsi’s blog.

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Code Optimization

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UPDATE: Thanks to @dietpepsi we further improved from 17ms to 15ms.

1. 59ms: Use search and startsWith in Trie class like this popular solution.
2. 33ms: Remove Trie class which unnecessarily starts from root in every dfs call.
3. 30ms: Use w.toCharArray() instead of w.charAt(i).
4. 22ms: Use StringBuilder instead of c1 + c2 + c3.
5. 20ms: Remove StringBuilder completely by storing word instead of boolean in TrieNode.
6. 20ms: Remove visited[m][n] completely by modifying board[i][j] = '#' directly.
7. 18ms: check validity, e.g., if(i > 0) dfs(...), before going to the next dfs.
8. 17ms: De-duplicate c - a with one variable i.
9. 15ms: Remove HashSet completely. dietpepsi’s idea is awesome.

The final run time is 15ms. Hope it helps!

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public List<String> findWords(char[][] board, String[] words) {
    List<String> res = new ArrayList<>();
    TrieNode root = buildTrie(words);
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            dfs (board, i, j, root, res);
        }
    }
    return res;
}

public void dfs(char[][] board, int i, int j, TrieNode p, List<String> res) {
    char c = board[i][j];
    if (c == '#' || p.next[c - 'a'] == null) return;
    p = p.next[c - 'a'];
    if (p.word != null) {   // found one
        res.add(p.word);
        p.word = null;     // de-duplicate
    }

    board[i][j] = '#';
    if (i > 0) dfs(board, i - 1, j ,p, res); 
    if (j > 0) dfs(board, i, j - 1, p, res);
    if (i < board.length - 1) dfs(board, i + 1, j, p, res); 
    if (j < board[0].length - 1) dfs(board, i, j + 1, p, res); 
    board[i][j] = c;
}

public TrieNode buildTrie(String[] words) {
    TrieNode root = new TrieNode();
    for (String w : words) {
        TrieNode p = root;
        for (char c : w.toCharArray()) {
            int i = c - 'a';
            if (p.next[i] == null) p.next[i] = new TrieNode();
            p = p.next[i];
       }
       p.word = w;
    }
    return root;
}

class TrieNode {
    TrieNode[] next = new TrieNode[26];
    String word;
}

原文鏈接：https://leetcode.com/problems/word-search-ii/discuss/59780/Java-15ms-Easiest-Solution-(100.00)