有意义的单词分割——经典dfs题目

 

680. 分割字符串

中文

English

给一个字符串,你可以选择在一个字符或两个相邻字符之后拆分字符串,使字符串由仅一个字符或两个字符组成,输出所有可能的结果

样例

样例1

输入： "123"
输出： [["1","2","3"],["12","3"],["1","23"]]

样例2

输入： "12345"
输出： [["1","23","45"],["12","3","45"],["12","34","5"],["1","2","3","45"],["1","2","34","5"],["1","23","4","5"],["12","3","4","5"],["1","2","3","4","5"]]

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

class Solution:     """     @param: : a string to be split     @return: all possible split string array     """     result = []           def splitString(self, s):         # write your code here         self.split_string_helper(s, start_index=0, path=[])         return self.result                     def split_string_helper(self, s, start_index, path):         if start_index >= len(s):             self.result.append(list(path))             return                   for step in (1, 2):             if start_index + step > len(s): # if no this clause, bug for "1" => ["1", "1"]                 return             comb = s[start_index:start_index+step]             path.append(comb)             self.split_string_helper(s, start_index+step, path)             path.pop()

 

582. 单词拆分II

中文

English

给一字串s和单词的字典dict,在字串中增加空格来构建一个句子，并且所有单词都来自字典。
返回所有有可能的句子。

样例

样例 1:

输入："lintcode"，["de","ding","co","code","lint"]
输出：["lint code", "lint co de"]
解释：
插入一个空格是"lint code"，插入两个空格是 "lint co de"

样例 2:

输入："a"，[]
输出：[]
解释：字典为空

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

import sys import json     def load_data():     data = json.loads(sys.stdin.read())     return data["string"], set(data["word_list"])     def dfs(s, start, words, arr, results):     if start == len(s):         results.append(",".join(arr))         return     for i in range(start, len(s)):         word = s[start:i+1]         if word in words:             arr.append(word)             dfs(s, i+1, words, arr, results)             arr.pop()     def main():     string, words = load_data()     results, arr = [], []     start = 0     dfs(string, start, words, arr, results)     results.sort()     print(json.dumps({'results': results}))     if __name__ == "__main__":     main()

当然，如果使用cache的话，更快：

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

class Solution:     """     @param: s: A string     @param: wordDict: A set of words.     @return: All possible sentences.     """       def wordBreak(self, s, wordDict):         # write your code here         self.cache = {}         return [" ".join(words) for words in self.dfs(s, words_dict=wordDict)]       def dfs(self, s, words_dict):         if s in self.cache:             return self.cache[s]           if not s:             return [[]]           result = []         for i in range(0, len(s)):             word = s[:i + 1]             if word in words_dict:                 for w in self.dfs(s[i + 1:], words_dict):                     result.append([word] + w)           self.cache[s] = result         return result