CareerCup String getSentence(String text, Set<String> dictionary); O(n)

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本文介绍了一种用于将无空格字符串根据字典高效拆分成独立词汇的算法，并提供了一个具体的Java实现示例。该算法利用递归方法遍历字符串的所有可能拆分方式，通过检查子串是否存在于给定的词汇表中来决定是否采用该拆分方案。

String getSentence(String text, Set<String> dictionary);

// text is a string without spaces, you need to insert spaces into text, so each word seperated by the space in the resulting string exists in the dictionary, return the resulting string

// running time has to be at least as good as O(n)

// getSentence("iamastudentfromwaterloo", {"from, "waterloo", "hi", "am", "yes", "i", "a", "student"}) -> "i am a student from waterloo"

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import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
	
	public static void main(String[] args) {
		
		Solution instance = new Solution();
		Set<String> dictionary = new HashSet();
		dictionary.add("google");
		dictionary.add("is");
		dictionary.add("awesome");
		List<String> store = new ArrayList<String>();
		instance.printWords("googleisawesome", store, dictionary);
		for(int i = store.size() - 1; i >= 0; --i) {
			System.out.println(store.get(i));
		}
	}


	private boolean printWords(String string, List<String> store, Set<String> dictionary) {
		if(string.length() == 0) {
			return true;
		}
		for(int i = 1; i <= string.length(); ++i) {
			String curWord = string.substring(0, i);
			if(dictionary.contains(curWord) && printWords(string.substring(i), store, dictionary)) {
				store.add(curWord);
				return true;
			}
		}
		return false;
	}
}

Trie tree is expected for this problem...