Solution_140.java

/*
140. 单词拆分 II
给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict，在字符串中增加空格来构建一个句子，使得句子中所有的单词都在词典中。
返回所有这些可能的句子。

说明：
分隔时可以重复使用字典中的单词。
你可以假设字典中没有重复的单词。

示例 1：
输入:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
输出:
[
  "cats and dog",
  "cat sand dog"
]

示例 2:
输入:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
输出:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
解释: 注意你可以重复使用字典中的单词。

示例 3：
输入:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
输出:
[]
 */

import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Solution solution = new Solution();
        String s = "catsanddog";
        //"catsanddog"
        String[] words = new String[]{"cat","cats","and","sand","dog"};
        List<String> wordDict = new ArrayList<>(Arrays.asList(words));
        System.out.println(solution.wordBreak(s, wordDict));
    }

    List<String> ans = new ArrayList<>();

    public List<String> wordBreak(String s, List<String> wordDict) {
        return solve(s, wordDict,0);
    }

    HashMap<Integer, List<String>> map = new HashMap<>();

    public List<String> solve(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 = solve(s, wordDict, end);
                for (String l : list) {
                    res.add(s.substring(start, end) + (l.equals("") ? "" : " ") + l);
                }
            }
        }
        map.put(start, res);
        return res;
    }

}