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| 1 | +class Node: |
| 2 | + def __init__(self, v, p=None, w=False): |
| 3 | + self.word = w #If the node represents the end of a word or not |
| 4 | + self.parent = p |
| 5 | + self.value = v |
| 6 | + self.children = {} |
| 7 | + |
| 8 | + |
| 9 | +class Trie: |
| 10 | + def __init__(self): |
| 11 | + self.root = Node('') #The root of the trie is always empty |
| 12 | + |
| 13 | + def Insert(self, word): |
| 14 | + """ |
| 15 | + Insert word in the trie. Starting from the root, move down the trie |
| 16 | + following the path of characters in the word. If the nodes for the word |
| 17 | + characters end, add them. When the last char is added, mark it as a |
| 18 | + word-ending node. |
| 19 | + """ |
| 20 | + l = len(word) |
| 21 | + curr = self.root |
| 22 | + for i, c in enumerate(word): |
| 23 | + last = False |
| 24 | + if(i == l-1): |
| 25 | + #The last char of the word |
| 26 | + last = True |
| 27 | + |
| 28 | + if(c not in curr.children): |
| 29 | + curr.children[c] = Node(c, curr, last) |
| 30 | + elif(last): |
| 31 | + #c already exists, but as it is the last char of word, |
| 32 | + #it should now be flagged as a word in the trie. |
| 33 | + curr.children[c].word = True |
| 34 | + |
| 35 | + curr = curr.children[c] |
| 36 | + |
| 37 | + def Search(self, word): |
| 38 | + """ |
| 39 | + Searches for given word in trie. We want to find the last node for the |
| 40 | + word. If we can't, then it means the word is not in the trie. |
| 41 | + """ |
| 42 | + if self.FindFinalNode(word): |
| 43 | + return True |
| 44 | + else: |
| 45 | + return False |
| 46 | + |
| 47 | + def FindWords(self, prefix): |
| 48 | + """ |
| 49 | + Find all words with the given prefix |
| 50 | + """ |
| 51 | + v = self.FindFinalNode(prefix) |
| 52 | + wList = self.BuildWordList(v, prefix) |
| 53 | + if(v and v.word): |
| 54 | + #v exists and the prefix is itself a word; add it to the list. |
| 55 | + wList.append(prefix) |
| 56 | + |
| 57 | + return wList |
| 58 | + |
| 59 | + def FindFinalNode(self, word): |
| 60 | + """ |
| 61 | + Returns the last node in given word. The process goes like this: |
| 62 | + Start from the root. For every char in word, go down one level. |
| 63 | + If we can't go down a level, then the word doesn't exist. |
| 64 | + If we do, and the current char is the last char of the word and |
| 65 | + the node we are currently at is a word, then we have found the given |
| 66 | + word. |
| 67 | + """ |
| 68 | + curr = self.root |
| 69 | + l = len(word) |
| 70 | + |
| 71 | + for i, c in enumerate(word): |
| 72 | + if(c not in curr.children): |
| 73 | + #There is no prefix of cWord + c |
| 74 | + return None |
| 75 | + |
| 76 | + if(i == l-1): |
| 77 | + #Last char of word |
| 78 | + return curr.children[c] |
| 79 | + |
| 80 | + curr = curr.children[c] |
| 81 | + |
| 82 | + return None |
| 83 | + |
| 84 | + def BuildWordList(self, v, cWord): |
| 85 | + """ |
| 86 | + Recursively builds the list of words. |
| 87 | + * v: Node to check |
| 88 | + * cWord : The word built up to v |
| 89 | + """ |
| 90 | + if(not v): |
| 91 | + return None |
| 92 | + |
| 93 | + wList = [] |
| 94 | + for i, k in v.children.items(): |
| 95 | + tempWord = cWord + i |
| 96 | + |
| 97 | + if(k.word): |
| 98 | + #If the iterated prefix is a word, add it to the list |
| 99 | + wList.append(tempWord) |
| 100 | + |
| 101 | + #The list of words under tWord |
| 102 | + wList.extend(self.BuildWordList(k, tempWord)) |
| 103 | + |
| 104 | + return wList |
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