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Problem Word Break

时间:2014-07-07 16:02:11      阅读:219      评论:0      收藏:0      [点我收藏+]

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Problem Description:

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

 

 Solution:
 1 public boolean wordBreak(String s, Set<String> dict) {
 2        Set<Character> set = new HashSet<Character>();
 3         for (int i = 0; i < s.length(); i++) {
 4             set.add(s.charAt(i));
 5         }
 6         for (char a : set) {
 7             boolean found = false;
 8             for (String word : dict) {
 9                 if (word.indexOf((int)a) >= 0) {
10                     found = true;
11                 }
12             }
13 
14             if (! found) {
15                 return false;
16             }
17         }
18 
19         if (s == null || dict.size() == 0 || dict == null) {
20             return false;
21         }
22         if (s.equals("")) return true;
23 
24         for (String word : dict) {
25             if (word.length() <= s.length()) {
26                 if (s.substring(0, word.length()).equals(word)) {
27                     boolean result = wordBreak(s.substring(word.length()), dict);
28                     if (result) return true;
29                 }
30             }
31         }
32 
33         return false;
34     }

 

Problem Word Break,布布扣,bubuko.com

Problem Word Break

标签:des   style   blog   color   strong   cti   

原文地址:http://www.cnblogs.com/liew/p/3815035.html

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