LeetCode 139. Word Break 动态规划DP Python解法

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。

这里写图片描述

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