題目如下：

Given a list of words (without duplicates), please write a program that returns all concatenated words in the given list of words.

A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.

Example:

Input: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats";
"dogcatsdog" can be concatenated by "dog", "cats" and "dog";
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

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Note:

The number of elements of the given array will not exceed?10,000

The length sum of elements in the given array will not exceed?600,000.

All the input string will only include lower case letters.

The returned elements order does not matter.

解題思路：我的方法是Input按單詞的長度升序排序后存入字典樹中，因為Input中單詞不重復，所有較長的單詞肯定是由比其短的單詞拼接而成的。每個單詞在存入字典樹的時候，同時做前綴匹配，并保存匹配的結果，接下來再對這些結果遞歸做前綴匹配，直到無法匹配或者完全匹配為止。如果有其中任意一個結果滿足完全匹配，則表示這個單詞可以由其他的單詞拼接而成。

代碼如下：

class TreeNode(object):def __init__(self, x):self.val = xself.childDic = {}self.hasWord = Falseclass Trie(object):dic = {}def __init__(self):"""Initialize your data structure here."""self.root = TreeNode(None)self.dic = {}def divide(self, word,flag):substr = []current = self.rootpath = ''for i in word:path += iif i not in current.childDic:current.childDic[i] = TreeNode(i)current = current.childDic[i]if current.hasWord:substr.append(word[len(path):])self.dic[path] = 1if flag:self.dic[word] = 1current.hasWord = Truereturn substrdef isExist(self,w):return w in self.dicclass Solution(object):def findAllConcatenatedWordsInADict(self, words):""":type words: List[str]:rtype: List[str]"""words.sort(cmp=lambda x1,x2:len(x1) - len(x2))t = Trie()res = []for w in words:queue = t.divide(w,True)while len(queue) > 0:item = queue.pop(0)if t.isExist(item):res.append(w)#t.dic[w] = 1breakelse:queue += t.divide(item,False)return res

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轉載于:https://www.cnblogs.com/seyjs/p/10508910.html

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