source: https://leetcode.com/problems/word-break/
Table of Contents
Word Break
Description
Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.
Note that the same word in the dictionary may be reused multiple times in the segmentation.
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
Constraints:
- 1 <= s.length <= 300
- 1 <= wordDict.length <= 1000
- 1 <= wordDict[i].length <= 20
- s and wordDict[i] consist of only lowercase English letters.
- All the strings of wordDict are unique.
Solution
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class Solution { public boolean wordBreak(String s, List<String> wordDict) { Set<String> set = new HashSet<>(wordDict); return helper(s, set, new HashMap<>()); } private boolean helper(String s, Set<String> wordDict, Map<String, Boolean> memo){ if(s.isEmpty()) return true; if(memo.containsKey(s)){ return memo.get(s); } boolean exist = false; for(int i=0;i<s.length();i++){ if(wordDict.contains(s.substring(0, i+1))){ exist = exist | helper(s.substring(i+1), wordDict, memo); } } memo.put(s, exist); return memo.get(s); } } |