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| 1 | +""" |
| 2 | +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. |
| 3 | + |
| 4 | +Note: |
| 5 | + |
| 6 | +The same word in the dictionary may be reused multiple times in the segmentation. |
| 7 | +You may assume the dictionary does not contain duplicate words. |
| 8 | +Example 1: |
| 9 | + |
| 10 | +Input: s = "leetcode", wordDict = ["leet", "code"] |
| 11 | +Output: true |
| 12 | +Explanation: Return true because "leetcode" can be segmented as "leet code". |
| 13 | +Example 2: |
| 14 | + |
| 15 | +Input: s = "applepenapple", wordDict = ["apple", "pen"] |
| 16 | +Output: true |
| 17 | +Explanation: Return true because "applepenapple" can be segmented as "apple pen apple". |
| 18 | + Note that you are allowed to reuse a dictionary word. |
| 19 | +Example 3: |
| 20 | + |
| 21 | +Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"] |
| 22 | +Output: false |
| 23 | + |
| 24 | +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。 |
| 25 | + |
| 26 | +思路: |
| 27 | +Dp: |
| 28 | +从0开始,若此分隔存在于给定字典中,则可以断开。 |
| 29 | + |
| 30 | +s = "leetcode", wordDict = ["leet", "code"] |
| 31 | + |
| 32 | +leetcode |
| 33 | + l e e t c o d e |
| 34 | +T F F F F F F F F |
| 35 | + |
| 36 | +leet |
| 37 | +s[0:0+4] in wordDict |
| 38 | + |
| 39 | +s[0+4] = True |
| 40 | + |
| 41 | + l e e t c o d e |
| 42 | +T F F F T F F F F |
| 43 | + 当搜索到这里时会再次进行重复的搜索。 |
| 44 | + |
| 45 | + |
| 46 | +--- |
| 47 | +emmm, 写法待改进。 |
| 48 | +这个写法思路一样,不过效率会低。 |
| 49 | + |
| 50 | +beat 3%. |
| 51 | + |
| 52 | +测试地址: |
| 53 | +https://leetcode.com/problems/word-break/description/ |
| 54 | + |
| 55 | +""" |
| 56 | +class Solution(object): |
| 57 | + def wordBreak(self, s, wordDict): |
| 58 | + """ |
| 59 | + :type s: str |
| 60 | + :type wordDict: List[str] |
| 61 | + :rtype: bool |
| 62 | + """ |
| 63 | + |
| 64 | + dp = [True] + [False] * len(s) |
| 65 | + |
| 66 | + for i in range(len(s)): |
| 67 | + for j in range(i+1): |
| 68 | + if dp[j] == True: |
| 69 | + for x in wordDict: |
| 70 | + if x == s[j:j+len(x)]: |
| 71 | + dp[j+len(x)] = True |
| 72 | + |
| 73 | + return dp[-1] |
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