LeetCode 139. Word Break 动态规划DP Python解法
2018-03-24 00:21
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题目
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.给定一个目标字符串和一组字符串,判断目标字符串能否拆分成数个字符串,这些字符串都在给定的那组字符串中。
For example, given
s = “leetcode”,
dict = [“leet”, “code”].
Return true because “leetcode” can be segmented as “leet code”.
思路
我们采用动态规划的方法解决,dp[i]表示字符串s[:i]能否拆分成符合要求的子字符串。我们可以看出,如果s[j:i]在给定的字符串组中,且dp[j]为True(即字符串s[:j]能够拆分成符合要求的子字符串),那么此时dp[i]也就为True了。按照这种递推关系,我们就可以判断目标字符串能否成功拆分。代码
def wordBreak(self, s, wordDict): len_dic = len(wordDict) len_str = len(s) dp = [0] * len_str for i in range(len_str): for j in range(len_dic): len_j = len(wordDict[j]) if i + 1 >= len_j and wordDict[j] == s[i + 1 - len_j:i + 1] and (i + 1 - len_j == 0 or dp[i - len_j] == 1): dp[i] = 1 break return dp[len_str - 1] == 1
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