LeetCode139 Word Break 单词切割

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

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题源：here；完整实现：here

思路：

两种方案，一种递归，一种动态规划。递归方法超时。

1 递归：

bool helper(string s, unordered_set<string> wordSet){
	if (s.empty()) return true;
	for (int i = 0; i < s.size(); i++){
		string profix = s.substr(0, i + 1);
		if (wordSet.find(profix) != wordSet.end()){
			string nextS = s;
			nextS.erase(0, i + 1);
			if (helper(nextS, wordSet))
				return true;
		}
	}

	return false;
}

bool wordBreak(string s, vector<string>& wordDict) {
	unordered_set<string> wordSet;
	for (string word : wordDict)
		wordSet.insert(word);
		
	return helper(s, wordSet);
}

2 动态规划

bool wordBreak2(string s, vector<string>& wordDict){
		unordered_set<string> wordSet;
		for (string word : wordDict)
			wordSet.insert(word);
		int sLen = s.size();
		vector<bool> record(sLen, false);
		for (int i = 0; i < sLen; i++){
			string subS = s.substr(0, i+1);
			if (wordSet.find(subS) != wordSet.end()){
				record[i] = true;
				continue;
			}
			for (int j = 1; j < i+1; j++){
				string subS = s.substr(i - j + 1, j);
				if (wordSet.find(subS) != wordSet.end())
					record[i] = record[i - j] || record[i];
			}
		}

		return record[sLen - 1];
}