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"]]

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

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：

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