[leetcode]Word Break II @ Python
原题地址:https://oj.leetcode.com/problems/word-break-ii/
题意:
Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
Return all such possible sentences.
For example,
given
s = "catsanddog"
,
dict = ["cat",
"cats", "and", "sand", "dog"]
.
A solution is ["cats and dog", "cat sand dog"]
.
解题思路:这道题不只像word break那样判断是否可以分割,而且要找到所有的分割方式,那么我们就要考虑dfs了。不过直接用dfs解题是不行的,为什么?因为决策树太大,如果全部遍历一遍,时间复杂度太高,无法通过oj。那么我们需要剪枝,如何来剪枝呢?使用word break题中的动态规划的结果,在dfs之前,先判定字符串是否可以被分割,如果不能被分割,直接跳过这一枝。实际上这道题是dp+dfs。
代码:
class Solution: # @param s, a string # @param dict, a set of string # @return a list of strings def check(self, s, dict): dp = [False for i in range(len(s)+1)] dp[0] = True for i in range(1, len(s)+1): for k in range(0, i): if dp[k] and s[k:i] in dict: dp[i] = True return dp[len(s)] def dfs(self, s, dict, stringlist): if self.check(s, dict): if len(s) == 0: Solution.res.append(stringlist[1:]) for i in range(1, len(s)+1): if s[:i] in dict: self.dfs(s[i:], dict, stringlist+‘ ‘+s[:i]) def wordBreak(self, s, dict): Solution.res = [] self.dfs(s, dict, ‘‘) return Solution.res
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