【leetcode】472. Concatenated Words

题目如下：

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".
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.

解题思路：我的方法是Input按单词的长度升序排序后存入字典树中，因为Input中单词不重复，所有较长的单词肯定是由比其短的单词拼接而成的。每个单词在存入字典树的时候，同时做前缀匹配，并保存匹配的结果，接下来再对这些结果递归做前缀匹配，直到无法匹配或者完全匹配为止。如果有其中任意一个结果满足完全匹配，则表示这个单词可以由其他的单词拼接而成。

代码如下：

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

转载于:https://www.cnblogs.com/seyjs/p/10508910.html

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