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Leetcode dfs Word Break II

2024-07-21 20:25:51   222人阅读


Word Break II

 Total Accepted: 15138 Total Submissions: 92228My Submissions

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"].



题意：
给定一个字符串和一个单词字典，找出可以用字典里的单词来分割这个字符串的各种组合。
思路：dfs 
void dfs(int start, string partition);
start表示当前要分割的字符串s[start:]，
设 k将 s分割为s[:k]和s[k:]，则 0 < k <= s.size() 且 s[:k]存在字典dict中
按上面这样的思路直接做会超时，可采用dfs + 剪枝的思路
增加一个bool数组 possible[k] 表示字符串s[k:]是否可用字典里的单词来分割


unordered_set<string> _dict;
vector<string> res;
vector<bool> possible;
string _s;
void dfs(int start, string partition){
	if(start >= _s.size()){
		res.push_back(partition.substr(1, partition.size() - 1));
		return;
	}
	for(int k = start + 1; k <= _s.size(); ++k){
		if(possible[k] &&  _dict.find(_s.substr(start, k  - start)) != _dict.end()){ //剪枝
			int size_before = res.size();
			dfs(k , partition + " " + _s.substr(start, k  - start));
			if(res.size() == size_before) possible[k] = false; //剪枝
		}
	}
}


vector<string> wordBreak(string s, unordered_set<string> &dict){
	_dict = dict;
	_s = s;
	possible = vector<bool>(s.size() + 1, true);
	dfs(0, "");
	return res;
}


Leetcode dfs Word Break II


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