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_140_WordBreakII.java
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_140_WordBreakII.java
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package leetcode_1To300;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _140_WordBreakII {
/**
* 140. Word Break II
* For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].
A solution is ["cats and dog", "cat sand dog"].
time : O(n^3)
space : O(n^3)
* @param s
* @param wordDict
* @return
*/
HashMap<Integer, List<String>> map = new HashMap<>();
// DFS
public List<String> wordBreak(String s, List<String> wordDict) {
return dfs(s, wordDict, 0);
}
public List<String> dfs(String s, List<String> wordDict, int start) {
if (map.containsKey(start)) {
return map.get(start);
}
List<String> res = new ArrayList<>();
if (start == s.length()) {
res.add("");
}
for (int end = start + 1; end <= s.length(); end++) {
if (wordDict.contains(s.substring(start, end))) {
List<String> list = dfs(s, wordDict, end);
for (String temp : list) {
res.add(s.substring(start, end) + (temp.equals("") ? "" : " ") + temp);
}
}
}
map.put(start, res);
return res;
}
}