word break 变形

/*
Given a string and dictionary of words, break the string into minimum number of words from the dictionary.
Ex:
{"jumped", "over", "some", "thing”, "something"}
"jumpedoversomething",
0000010002000300003

Should return { “jumped”, “over”, “something”}

*/

 

fb: 求最少几次 直接return can[length]

class Solution {
 
  public static void main(String[] args) {
    HashSet<String> set = new HashSet<String>();
    set.add("jumped");
    set.add("over");
    set.add("some");
    set.add("thing");
    set.add("something");
    
    String target = "jumpedoversomething";
    List<String> res =wordBreak(target, set);
    for (String ss : res) {
        System.out.println(ss);
    }
    
  }
  
  
  public static List<String> wordBreak(String s, HashSet<String> wordDict) {
         int[] can = new int[s.length() + 1];
        HashMap<Integer, Integer> map = new HashMap<>();
        int max = maxWordLength(wordDict);
        // function
        can[0] = 1;
        for(int i = 1; i <= s.length(); i++) {
           
            for (int j = 1; j <= max && j <= i; j++) {
                if (can[i - j] == 0) {
                    continue;
                }
                String sub = s.substring(i - j, i);
                if (wordDict.contains(sub)) {
                    if (can[i] == 0) {
                        can[i] =  can[i - j] + 1;
                        map.put(i, i - j);
                    } else if (can[i - j] + 1 < can[i]) {
                        can[i] =  can[i - j] + 1;
                        map.put(i, i - j);
                    }
                   
                }
            }
        }
        List<String> ans = new ArrayList<>();
        int i = s.length(), j = 0;
        for (int k = 0; k < can[s.length()] - 1; k++) {
            j = (int) map.get(i);
            
            ans.add(s.substring(j, i));
            i = j;
        }
        return ans;
    }
     
    private static int maxWordLength(Set<String> wordDict) {
        int max = 0;
        for (String word: wordDict) {
            max = Math.max(max, word.length());
        }
        return max;
    }
}