LeetCode : 139. Word Break 拆分字符串

以你之姓@ 2021-06-24 16:11 389阅读 0赞

**试题**  
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.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.  
You may assume the dictionary does not contain duplicate words.  
Example 1:

Input: s = “leetcode”, wordDict = \[“leet”, “code”\]  
Output: true  
Explanation: Return true because “leetcode” can be segmented as “leet code”.  
Example 2:

Input: s = “applepenapple”, wordDict = \[“apple”, “pen”\]  
Output: true  
Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”.  
Note that you are allowed to reuse a dictionary word.  
Example 3:

Input: s = “catsandog”, wordDict = \[“cats”, “dog”, “sand”, “and”, “cat”\]  
Output: false

**代码**  
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class Solution {
        public boolean wordBreak(String s, List<String> wordDict) {
            HashSet<String> dict = new HashSet<>();
            HashSet<Integer> flag = new HashSet<>();
            for(String d : wordDict){
                dict.add(d);
            }
            return split(s,0,dict,flag);
        }
        public boolean split(String s, int start, HashSet<String> dict, HashSet<Integer> flag){
            if(start==s.length()){
                return true;
            }
            if(flag.contains(start)){
                return false;
            }
            for(int i=start+1; i<=s.length(); i++){
                String l = s.substring(start,i);
                if( dict.contains(l) && split(s,i,dict,flag) )
                {
                    return true;
                }
            }
            flag.add(start);
            return false;
        }
    }

超时代码

class Solution {
        public boolean wordBreak(String s, List<String> wordDict) {
            HashSet<String> dict = new HashSet<>();
            for(String d : wordDict){
                dict.add(d);
            }
            return split(s,dict);
        }
        public boolean split(String s, HashSet<String> dict){
            if(dict.contains(s)){
                return true;
            }
            if(s.length()==1){
                return false;
            }
            for(int i=1; i<s.length(); i++){
                String l = s.substring(0,i);
                String r = s.substring(i,s.length());
                if( (dict.contains(l)||split(l,dict)) && (dict.contains(r)||split(r,dict)) ){
                    return true;
                }
            }
            return false;
        }
    }