1+ // Source : https://leetcode.com/problems/word-break/
2+ // Author : henrytien
3+ // Date : 2022年02月02日
4+ 5+ /* ****************************************************************************************************
6+ *
7+ * Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a
8+ * space-separated sequence of one or more dictionary words.
9+ *
10+ * Note that the same word in the dictionary may be reused multiple times in the segmentation.
11+ *
12+ * Example 1:
13+ *
14+ * Input: s = "leetcode", wordDict = ["leet","code"]
15+ * Output: true
16+ * Explanation: Return true because "leetcode" can be segmented as "leet code".
17+ *
18+ * Example 2:
19+ *
20+ * Input: s = "applepenapple", wordDict = ["apple","pen"]
21+ * Output: true
22+ * Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
23+ * Note that you are allowed to reuse a dictionary word.
24+ *
25+ * Example 3:
26+ *
27+ * Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
28+ * Output: false
29+ *
30+ * Constraints:
31+ *
32+ * 1 <= s.length <= 300
33+ * 1 <= wordDict.length <= 1000
34+ * 1 <= wordDict[i].length <= 20
35+ * s and wordDict[i] consist of only lowercase English letters.
36+ * All the strings of wordDict are unique.
37+ ******************************************************************************************************/
38+ 39+ #include " ../inc/ac.h"
40+ 41+ class Solution
42+ {
43+ public:
44+ bool wordBreak (string s, vector<string> &wordDict)
45+ {
46+ if (s.empty () || wordDict.size () == 0 )
47+ {
48+ return false ;
49+ }
50+ 51+ vector<bool > dp (s.size ()+1 ,false );
52+ set<string> dict;
53+ for (auto &&w : wordDict)
54+ {
55+ dict.insert (w);
56+ }
57+ 58+ dp[0 ] = true ;
59+ 60+ for (int i = 1 ; i <= s.size (); i++)
61+ {
62+ for (int j = i - 1 ; j >= 0 ; j--)
63+ {
64+ if (dp[j])
65+ {
66+ string word = s.substr (j,i- j);
67+ if (dict.find (word) != dict.end ())
68+ {
69+ dp[i] = true ;
70+ break ;
71+ }
72+ 73+ }
74+ }
75+ }
76+ return dp[s.size ()];
77+ 78+ }
79+ };
80+ 81+ int main ()
82+ {
83+ string s = " applepenapple"
84+ vector<string> word_dict{" apple" " pen"
85+ cout << Solution ().wordBreak (s, word_dict) << " \n "
86+ return 0 ;
87+ }
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