Word Break II

最新推荐文章于 2025-08-11 14:43:21 发布

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本文介绍了一种使用深度优先搜索(DFS)算法解决字符串拆分问题的方法，并通过加入备忘录来优化性能。该问题要求将给定字符串通过字典中的词汇进行拆分，并返回所有可能的拆分组合。

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题目描述：

Given a string s and a dictionary of words dict, add spaces ins to construct a sentence where each word is a valid dictionary word.

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"].

dfs会超时：

public class Solution {
    public List<String> wordBreak(String s, Set<String> wordDict) {
		List<String> result=new ArrayList<String>();
		StringBuilder sb=new StringBuilder();
		getScentence(s, wordDict, 0, sb, result);
		return result;
    }
	
	public void getScentence(String s, Set<String> wordDict,int start,StringBuilder sb,List<String> result){
		if(start==s.length()){
			int len=sb.length();
			String str=sb.delete(len-1, len).toString();
			result.add(str);return;
		}
		int len=sb.length();
		for(int i=start+1;i<=s.length();i++){
			String substr=s.substring(start, i);
			if(wordDict.contains(substr)){
				sb.append(substr).append(" ");
				getScentence(s, wordDict, i, sb, result);
				sb.delete(len, len+substr.length()+1);//这里要+1，小心细节- -！
			}
		}
	}
}

带备忘录的dfs算法：

public class Solution {
    Map<String, List<String>> mem = new HashMap<>();
	public List<String> wordBreak(String s, Set<String> wordDict) {
	    if (mem.get(s) != null) return mem.get(s);
	    List<String> result = new ArrayList<>();
	    if (wordDict.contains(s)) result.add(s);
	    for (int i = 1; i < s.length(); i++) {
	        String word = s.substring(0, i);
	        if (wordDict.contains(word)) {
	            List<String> tmp = wordBreak(s.substring(i), wordDict);
	            for (String str : tmp) {
	                result.add(word + " " + str);
	            }
	        }
	    }
	    mem.put(s, result);
	    return result;
	}
}