139. Word Break -Medium

Question

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

给出一个非空字符串s和一个包含一系列非空单词的字典wordDict，请你确定是否s可以用空格分割出一个或多个字典中的单词。字典中的单词不重复。

Example

*For example, given

s = “leetcode”,

dict = [“leet”, “code”].

Return true because “leetcode” can be segmented as “leet code”.

Solution

• 动态规划解。定义dp[i]：s[:i]是否在字典中。那么递推式怎么找呢？假设我们要判断dp[i]是否在字典中，如果我们已知dp[j]在字典中（j < i），那么如果我们又判断得到s[j+1:i]也在字典中，那么我们就可以知道dp[i]在字典中了。即 dp[i] = dp[j] and s[j+1:i] in wordDict (j < i)

  class Solution(object):
      def wordBreak(self, s, wordDict):
          """
          :type s: str
          :type wordDict: List[str]
          :rtype: bool
          """
          dp = [True] + [False] * len(s)
          for index_s in range(1, len(s) + 1):
              for j in range(index_s):
                  # 如果前j个已经可以在字典中找到且j+1:index_s也在字典中（因为index的范围是1 -- len(s) + 1，所以写j:index_s）
                  if dp[j] and s[j:index_s] in wordDict:
                          # 代表前index_s个字符串可以在字典中找到（包括分解找到）
                          dp[index_s] = True
                          break
          return dp[-1]