Word Break

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".

 

 1 public class Solution {
 2     public boolean wordBreak(String s, Set<String> dict) {
 3         // Note: The Solution object is instantiated only once and is reused by each test case.
 4         if(s == null || s.length() == 0) return true;
 5         int len = s.length();
 6         boolean[] map = new boolean[len];
 7         for(int i = len - 1; i > -1; i --){
 8             for(int j = i + 1; j <= len; j ++){
 9                 String ss = s.substring(i,j);
10                 if(dict.contains(ss) && j == len){
11                     map[i] = true;
12                     break;
13                 }
14                 else if(dict.contains(ss) && j < len && map[j] == true){
15                     map[i] = true;
16                     break;
17                 }
18             }
19         }
20         return map[0];
21     }
22 }

 第二遍：

 1 public class Solution {
 2     public boolean wordBreak(String s, Set<String> dict) {
 3         // Note: The Solution object is instantiated only once and is reused by each test case.
 4         if(s == null || s.length() == 0) return true;
 5         int len = s.length();
 6         boolean[] map = new boolean[len];
 7         for(int i = len - 1; i > -1; i --)
 8             for(int j = i + 1; j <= len; j ++){
 9                     if(j < len && !map[j]) continue;
10                     else if(dict.contains(s.substring(i, j)) && (j == len || map[j])){
11                         map[i] = true;
12                         break;
13                     }
14         }
15         return map[0];
16     }
17 }

 

posted on 2013-10-09 11:32  Step-BY-Step  阅读(223)  评论(0)  编辑  收藏  举报

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