标签：c style class blog code java
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
[leetcode]Word Break II @ Python,布布扣,bubuko.com
[leetcode]Word Break II @ Python
标签：c style class blog code java
原文地址：http://www.cnblogs.com/zuoyuan/p/3760804.html
