【leetcode】(python)140. Word Break II单词分割II

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给定一个非空字符串s和一个包含非空单词的字典wordDict，构造所有可能将字符串s分割成合法单词序列的方法。注意字典中的单词可以重复使用，且假设字典中不含有重复单词。

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单词分割II

Description
Example
题意
解题思路
- code

140. Word Break II Hard

Description

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.

Example

Example 1:

Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

题意

求非空字符串s的单词分割的所有可能结果。

解题思路

这题在139. Word Break词语分割的基础上求所有可能结果。一般求最优用动态规划，求所有可能值用回溯，也即递归。这题常规递归时间复杂度极大超时，所以采用了字典记录已分割单词的方式能求得结果。

code

class Solution(object):
    def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: List[str]
        :rtype: List[str]
        """
        memo = dict()
        return self.dfs(s, wordDict,memo)
                
    def dfs(self, s, wordDict, memo):
        if not s:
            return []
        if s in memo:
            return memo[s]
        res = []
        if s in wordDict:
            res.append(s)
        for i in range(1,len(s)):
            if s[:i] in wordDict:
                for j in self.dfs(s[i:],wordDict,memo):
                    res.append(s[:i] + " " + j)
        memo[s] = res
        return res