131. Palindrome Partitioning and 140. Word Break II

Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

[
  ["aa","b"],
  ["a","a","b"]
]

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<List<String>>();
        if(s == null || s.length() == 0) return res;
        dfs(s, res, new ArrayList<String>(), 0);
        return res;
    }
    
    public void dfs(String s, List<List<String>> res, List<String> path, int pos){
        if(pos == s.length()){
            res.add(new ArrayList<String>(path));
            return;
        }
        
        for(int i = pos; i < s.length(); i++){
            // 找到开头的某个palindrome进行切割。
            // 剩下的部分就是相同的子问题。
            if(isPalindrome(s, pos, i)){
                path.add(s.substring(pos, i+1));
                dfs(s, res, path, i+1);
                path.remove(path.size()-1);
            }
        }
    }
    
    public boolean isPalindrome(String s, int i, int j){
        while(i < j){
            if(s.charAt(i++) != s.charAt(j--)) return false;
        }
        return true;
    }
}

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

public class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        return dfs(s, wordDict, new HashMap<String, List<String>>());
    }
    
    public List<String> dfs(String s, List<String> wordDict, HashMap<String, List<String>> memo){
        // 记忆化搜索，可以减少重复部分的操作，直接得到break后的结果。
        if(memo.containsKey(s)){
            return memo.get(s);
        }
        
        int n = s.length();
        List<String> list = new ArrayList<String>();
        for(String word: wordDict){
            // 如果这个单词可以作为开头，把剩下的部分作为完全相同的子问题。
            // 得到的结果和这个单词组合在一起得到结果。
            if(!s.startsWith(word)) continue;
            
            int len = word.length();
            if(len == n){
                list.add(word);
            } else {
                List<String> sublist = dfs(s.substring(len), wordDict, memo);
                for(String item: sublist){
                    list.add(word + " " + item);
                }
            }
        }
        
        memo.put(s, list);
        return list;
    }
}