Leetcode139 WordBreak解题方法

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本文详细解析了LeetCode139WordBreak题目，介绍了一种使用HashSet存储字典和HashMap存储记忆的解决方案，避免重复计算，提高效率。

Leetcode139 WordBreak

题目描述

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

思路

思路：使用HashSet存储字典，使用HashMap存储记忆，如果left和right之前进行过匹配，则直接调用之前的记忆，无需再花时间循环。

代码

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        HashMap<String, Boolean> m = new HashMap<>();
        Set<String> dict = new HashSet<String>(wordDict);
        return wordBreak(s, dict,m);
    }
    boolean wordBreak(String s, Set<String> wordDict, HashMap<String,Boolean> m){
        if(m.containsKey(s)) return m.get(s);
        if(wordDict.contains(s)){
            m.put(s,true);
            return true;
        }
        for(int i=1;i<s.length();i++){
            String left = s.substring(0,i);
            String right = s.substring(i);
            if(wordDict.contains(right)&&wordBreak(left,wordDict,m)){
                m.put(s,true);
                return true;
            }
        }
        m.put(s,false);
        return false;
    }
}