Trie

算法思路：

Leetcode 208 Implement Trie (Prefix Tree), 也可以用HashMap将所有前缀加入到Map来实现，效率稍低。

注意事项：

1. TrieNode用{}和is_end，insert, search, startswith用it和i迭代比较
2. startswith含整个单词，如单词apple，startswith(‘apple’) -> True
3. Line 11记得加，也就是dict在取value是一定要先检验key是否存在。可以不加，解决方案是用defaultdict（后版本）。

新版本比旧版本更简洁，体现在is_end的处理放在了for循环外。
存储结构举例： a

TrieNode(
  children = { 
              'a':  TrieNode(is_End = True)
		     }
  is_end = False
)

Python代码：

    def __init__(self):
        self.head = TrieNode()

    def insert(self, word: str) -> None:
        it = self.head
        for i in range(len(word)):
            # if word[i] not in it.children:
                #it.children[word[i]] = TrieNode()
            it = it.children[word[i]]
        it.is_end = True

    def search(self, word: str) -> bool:
        it = self.head
        for i in range(len(word)):
            if word[i] not in it.children:
                return False
            it = it.children[word[i]]
        return it.is_end

    def startsWith(self, prefix: str) -> bool:
        it = self.head
        for i in range(len(prefix)):
            if prefix[i] not in it.children:
                return False
            it = it.children[prefix[i]]
        return True

class TrieNode:

    def __init__(self):
        self.children = collections.defaultdict(TrieNode)  # {}
        self.is_end = False

旧版本： 冗余了一个if语句if i == len(word) - 1

Python代码：

class Trie(TestCases):

        def __init__(self):
        self.head = TrieNode()

    def insert(self, word: str) -> None:
        if not word:
            return
        it = self.head
        for i in range(len(word)):
            # if word[i] not in it.children:
                # it.children[word[i]] = TrieNode()
            it = it.children[word[i]]
            if i == len(word) - 1:
                it.is_end = True

    def search(self, word: str) -> bool:
        if not word:
            return False
        it = self.head
        for i in range(len(word)):
            if word[i] not in it.children:
                return False
            it = it.children[word[i]]
            if i == len(word) - 1 and it.is_end:
                return True
        return False

    def startsWith(self, prefix: str) -> bool:
        if not prefix:
            return False
        it = self.head
        for i in range(len(prefix)):
            if prefix[i] not in it.children:
                return False
            it = it.children[prefix[i]]
            if i == len(prefix) - 1:
                return True
        return False

class TrieNode:

    def __init__(self):
        self.children = collections.defaultdict(TrieNode)  # {}
        self.is_end = False

算法分析：

每个操作时间复杂度为O(n)，空间复杂度O(n)，n为单词长度。