leetcode 139. Word Break
2017-06-11 16:14
375 查看
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. You may assume the dictionary does not contain duplicate words. For example, given s = "leetcode", dict = ["leet", "code"]. Return true because "leetcode" can be segmented as "leet code". UPDATE (2017/1/4): The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.
采用动态规划,dp[i]表示目标字符串s中0-i是否合法
public class Solution { public boolean wordBreak(String s, List<String> wordDict) { boolean[] res = new boolean[s.length() +1]; res[0] = true; for(int i=1; i < s.length()+1; i++){ for(String str : wordDict){ if(str.length() <=i){ if(res[i-str.length()]){ if(s.substring(i-str.length(), i).equals(str)){ res[i] = true; break; } } } } } return res[s.length()]; } }
相关文章推荐
- LeetCode 139 Word Break (DFS 分词 解存在性判断)
- 【LeetCode】139 - Word Break
- LeetCode-139. Word Break (JAVA)单词切分
- LeetCode 139. Word Break
- [leetcode]139. Word Break
- LeetCode-139. Word Break
- LeetCode - 139/140 - Word Break
- Leetcode 139 Word Break
- LeetCode(139)Word Break
- leetcode 139. Word Break
- leetcode 139. Word Break
- leetcode 139 —— Word Break
- Leetcode-139. Word Break
- Leetcode—139. Word Break
- [leetcode] 139.Word Break
- [LeetCode]139 Word Break
- [Leetcode 139, medium] Word Break
- leetcode 139. Word Break
- Leetcode 139 Word Break
- LeetCode 139 Word Break