[LeetCode]Word BreakII

最新推荐文章于 2025-08-08 21:26:37 发布

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本文介绍了一种使用深度优先搜索(DFS)解决字符串拆分问题的方法，通过预先构建一个标记矩阵来快速判断子串是否存在于词典中，进而找出所有可能的拆分方式。

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Given a string s and a dictionary of words dict, add spaces in s 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"].

Analysis:

For the "Return all " problems, usually DFS or BFS will work well.

In this problem, a DFS with a simple cutting edge condition will pass all the test cases.

The idea is to use mp[][] array to record which substring is in dict. and a array to record the insert index

for those substrings, we use DFS.

c++

class Solution {
public:
    void dfsWordBreak(string &s,
                  int st,
                  vector<int>&sp,
                  unordered_set<string> &dict,
                  vector<string> &result,
                  vector<vector<bool>> &mp){
    if(st>=s.size()){
        string str = s;
        for(int i=0;i<sp.size()-1;i++){
            str.insert(sp[i]+i," ");
        }
        result.push_back(str);
    }else{
        for(int j=0;j<mp[st].size();j++){
            if(mp[st][j]==true){
                sp.push_back(j+1);
                dfsWordBreak(s,j+1,sp,dict,result,mp);
                sp.pop_back();
            }
        }
    }
}
vector<string> wordBreak(string s, unordered_set<string> &dict) {
    vector<string> result;
    vector<vector<bool>> mp(s.size(),vector<bool>(s.size(),false));
    for(int i=0;i<s.size();i++){
        for(int j=i;j<s.size();j++){
            if(dict.find(s.substr(i,j-i+1))!=dict.end())
                mp[i][j] = true;
        }
    }
    bool flag = false;
    for(int i=0;i<s.size();i++){
        if(mp[i][s.size()-1])
            {flag = true;
            break;}
    }
    if(!flag) return result;
    vector<int>sp;
    dfsWordBreak(s,0,sp,dict,result,mp);
    return result;
}
};

Java

public class Solution {
    List<String> result;
	List<Integer> solu;
	boolean [][] mp;
	public List<String> wordBreak(String s, Set<String> dict) {
        result = new ArrayList<>();
        solu = new ArrayList<>();
        mp = new boolean[s.length()][s.length()];
        for(int i=0;i<s.length();i++){
        	for(int j=i;j<s.length();j++){
        		if(dict.contains(s.substring(i, j+1)))
        			mp[i][j] = true;
        	}
        }
        boolean flag = false;
        for(int i=0;i<s.length();i++){
        	if(mp[i][s.length()-1]){
        		flag = true;
        		break;
        	}
        }
        if(!flag) return result;
        dfs(s, 0, dict);
        return result;
    }
	public void dfs(String s, int st, Set<String> dict ){
		if(st>=s.length()){
			StringBuffer str = new StringBuffer(s);
			for(int i=0;i<solu.size()-1;i++){
				str.insert(solu.get(i)+i, " ");
			}
			result.add(str.toString());
		}else {
			for(int j=0;j<mp[st].length;j++){
				if(mp[st][j]==true){
					solu.add(j+1);
					dfs(s, j+1, dict);
					solu.remove(solu.size()-1);
				}
			}
		}
	}
}