Java Word Break(单词拆解)

给定一个字符串 String s = "leetcode"
dict = ["leet", "code"].

查看一下是够是字典中的词语组成，如果是返回true，否则返回false。

下边提供3种思路

1.动态算法

import java.util.HashSet;import java.util.Set;public class WordBreak1 {public static void main(String[] args) {//"["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]//String s="aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab";String s ="LeetCodea";Set<String> dict = new HashSet<String>();dict.add("Leet");dict.add("Code");dict.add("a");System.out.println(wordBreak(s,dict));}public static boolean wordBreak(String s, Set<String> dict) {boolean[] t = new boolean[s.length() + 1];t[0] = true; // set first to be true, why?// Because we need initial statefor (int i = 0; i < s.length(); i++) {// should continue from match positionif (!t[i])continue;for (String a : dict) {int len = a.length();int end = i + len;if (end > s.length())continue;if (t[end])continue;if (s.substring(i, end).equals(a)) {t[end] = true;}}}return t[s.length()];}}

2.普通算法(1)

import java.util.Set;public class WorkBreak2 {public boolean wordBreak(String s, Set<String> dict) {return wordBreakHelper(s, dict, 0);}public boolean wordBreakHelper(String s, Set<String> dict, int start) {if (start == s.length())return true;for (String a : dict) {int len = a.length();int end = start + len;// end index should be <= string lengthif (end > s.length())continue;if (s.substring(start, start + len).equals(a))if (wordBreakHelper(s, dict, start + len))return true;}return false;}}

3.普通算法(2)

import java.util.Set;public class WordBreak3 {public static boolean wordBreak(String s, Set<String> dict) {// input validation// Base caseif (dict.contains(s))return true;else {for (int i = 0; i < s.length(); i++) {String sstr = s.substring(0, i);if (dict.contains(sstr))return wordBreak(s.substring(i), dict);}}return false;}}

但是以上的算法有一个问题，就是遇到这种情况，INPUT: "programcreek", ["programcree","program","creek"]. 无能为力。

大家讨论下吧？