Leetcode - Implement Trie (Prefix Tree)
Leetcode - Word Search

Leetcode - Word Search II

violet posted @ Mar 27, 2020 10:51:31 AM in 算法 with tags Algorithm Trie Golang backtracking , 45 阅读

https://leetcode.com/problems/word-search-ii/

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

 

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"]

 

func findWords(board [][]byte, words []string) []string {
    if len(board) == 0 || len(board[0]) == 0 {
        return []string{}
    }
    root := buildTrie(words)
    result := []string{}
    for i := 0; i < len(board); i++ {
        for j := 0; j < len(board[0]); j++ {
            dfs(board, root, &result,  i, j)
        }
    }
    return result
}

func dfs(board [][]byte, p *Trie, result *[]string, i, j int) {
    if p == nil {
        return
    }
    directions := [][]int{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}
    c := board[i][j]
    if c == '#' {
        return
    }
    p = p.Children[c - 'a']
    if p != nil && p.Val != "" {
        *result = append(*result, p.Val)
        p.Val = ""
    }
    board[i][j] = '#'
    for _, d := range directions {
        newx := i + d[0]
        newy := j + d[1]
        if newx >= 0 && newy >= 0 && newx < len(board) && newy < len(board[0]) {
            dfs(board, p, result, newx, newy)
        }
    }
    board[i][j] = c
}

type Trie struct {
    Val string
    Children []*Trie
}

func buildTrie(words []string) *Trie {
    root := &Trie{Children: make([]*Trie, 26)}
    for _, word := range words {
        node := root
        for i := 0; i < len(word); i++ {
            index := word[i] - 'a'
            if node.Children[index] == nil {
                node.Children[index] = &Trie{
                    Children: make([]*Trie, 26),
                }
            }
            node = node.Children[index]
        }
        node.Val = word
    }
    return root
}

func search(root *Trie, prefix []byte) *Trie {
    node := root
    for i := 0; i < len(prefix); i++ {
        index := prefix[i] - 'a'
        if node.Children[index] == nil {
            return nil
        }
        node = node.Children[index]
    }
    return node
}

func contains(arr []string, str string) bool {
    for _, s := range arr {
        if s == str {
            return true
        }
    }
    return false
}

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