word break medium

Author: weiy

DP/WordBreak.py

@@ -0,0 +1,73 @@
+"""
+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.
+
+Note:
+
+The same word in the dictionary may be reused multiple times in the segmentation.
+You may assume the dictionary does not contain duplicate words.
+Example 1:
+
+Input: s = "leetcode", wordDict = ["leet", "code"]
+Output: true
+Explanation: Return true because "leetcode" can be segmented as "leet code".
+Example 2:
+
+Input: s = "applepenapple", wordDict = ["apple", "pen"]
+Output: true
+Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
+             Note that you are allowed to reuse a dictionary word.
+Example 3:
+
+Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
+Output: false
+
+给一个非空字符串和一个包含单词的非空字典。判断是否能用字典里的单词组合成给定的字符串。
+
+思路：
+Dp:
+从0开始，若此分隔存在于给定字典中，则可以断开。
+
+s = "leetcode", wordDict = ["leet", "code"]
+
+leetcode
+  l e e t c o d e 
+T F F F F F F F F 
+
+leet
+s[0:0+4] in wordDict
+
+s[0+4] = True
+
+  l e e t c o d e 
+T F F F T F F F F 
+        当搜索到这里时会再次进行重复的搜索。
+
+
+---
+emmm, 写法待改进。
+这个写法思路一样，不过效率会低。
+
+beat 3%.
+
+测试地址：
+https://leetcode.com/problems/word-break/description/
+
+"""
+class Solution(object):
+    def wordBreak(self, s, wordDict):
+        """
+        :type s: str
+        :type wordDict: List[str]
+        :rtype: bool
+        """
+            
+        dp = [True] + [False] * len(s)
+
+        for i in range(len(s)):
+            for j in range(i+1):
+                if dp[j] == True:
+                    for x in wordDict:
+                        if x == s[j:j+len(x)]:
+                            dp[j+len(x)] = True 
+
+        return dp[-1]