Leetcode 212 Python

最新推荐文章于 2019-08-20 19:15:23 发布

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最新推荐文章于 2019-08-20 19:15:23 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/u010429989/article/details/79297399

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本文介绍了一种利用 Trie 树和深度优先搜索 (DFS) 的算法来解决在矩阵中查找字典词的问题。该算法首先将字典中的单词插入到 Trie 树中，然后遍历矩阵中的每个元素，通过 DFS 查找是否存在匹配的单词。

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解题思路：Trie+dfs，先将word存入trie，然后dfs遍历board，如果前缀不在word则退出，如果是word，且ans中没有，则加入ans。注：trie实现时需用map，否则tle

class TrieNode:
    # Initialize your data structure here.
    def __init__(self):
        self.children = dict()
        self.isWord = False
class Trie:
    def __init__(self):
        self.root = TrieNode()
    def insert(self, word):
        node = self.root
        for letter in word:
            child = node.children.get(letter)
            if child is None:
                child = TrieNode()
                node.children[letter] = child
            node = child
        node.isWord = True
    def search(self, word):
        node = self.root
        for letter in word:
            node = node.children.get(letter)
            if node is None:
                return False
        return node.isWord
    def startsWith(self, prefix):
        node = self.root
        for letter in prefix:
            node = node.children.get(letter)
            if node is None:
                return False
        return True

class Solution(object):
    def findWords(self, board, words):
        """
        :type board: List[List[str]]
        :type words: List[str]
        :rtype: List[str]
        """
        m = len(board)
        n = len(board[0])
        trie = Trie()
        for w in words:
            trie.insert(w)
        ans = []
        for i in range(m):
            for j in range(n):
                tmp = [[0 for q in range(n)] for q in range(m)]
                str = board[i][j]
                self.findWord(i,j,trie,board,tmp,str,ans)   
        return ans
    def findWord(self,i,j,trie,board,tmp,str,ans):
        m = len(board)
        n = len(board[0])
        tmp[i][j] = 1
        if trie.startsWith(str) == False:
            tmp[i][j] = 0
            return
        if trie.search(str) and str not in ans:
            ans.append(str)
        if i>0 and tmp[i-1][j] == 0:
            self.findWord(i-1,j,trie,board,tmp,str+board[i-1][j],ans)
        if j>0 and tmp[i][j-1] == 0:
            self.findWord(i,j-1,trie,board,tmp,str+board[i][j-1],ans)
        if i<m-1 and tmp[i+1][j] == 0:
            self.findWord(i+1,j,trie,board,tmp,str+board[i+1][j],ans)
        if j<n-1 and tmp[i][j+1] == 0:
            self.findWord(i,j+1,trie,board,tmp,str+board[i][j+1],ans)
        tmp[i][j] = 0