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。假设是返回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 state for (int i = 0; i < s.length(); i++) { // should continue from match position if (!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 length if (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 case if (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; } }
可是以上的算法有一个问题,就是遇到这样的情况。INPUT: "programcreek", ["programcree","program","creek"]. 无能为力。
大家讨论下吧?
时间: 2024-11-03 20:56:27