leetcode -- 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".

[解题思路]

1.brute force will TLE.

just check every substring from index 0 to the end.

 1 public boolean wordBreak(String s, Set<String> dict) {
 2         // Note: The Solution object is instantiated only once and is reused by each test case.
 3         int len = s.length();
 4         boolean flag = false;
 5               
 6         for(int i = 1; i <= len; i++){
 7             String subStr = s.substring(0, i);
 8             if(dict.contains(subStr)){
 9                 if(subStr.length() == s.length()){
10                     return true;
11                 }   
12                 flag = wordBreak(s.substring(i), dict);
13             }
14         }
15         return flag;
16     }

2.DP

Reference the dicussion in leetcode.

Here we use seg(i, j) to demonstrate whether substring start from i and length is j is in dict?

base case:

when j = 0; seg(i, j) = false;

State transform equation:

seg(i, j) = true. if s.substring(i, j - 1) is in dict.

else seg(i, j) = seg(i, k) && seg(i + k, j - k);

 1 public boolean wordBreak(String s, Set<String> dict) {
 2         // Note: The Solution object is instantiated only once and is reused by each test case.
 3         if(s == null || dict.size() <= 0){
 4             return false;
 5         }
 6         
 7         int length = s.length();
 8         // seg(i, j) means substring t start from i and length is j can be segmented into 
 9         // dictionary words
10         boolean[][] seg = new boolean[length][length + 1];
11         for(int len = 1; len <= length; len++){
12             for(int i = 0; i < length; i++){
13                 String t = s.substring(i, i + len);
14                 if(dict.contains(t)){
15                     seg[i][len] = true;
16                     continue;
17                 }
18                 for(int k = 1; k < len; k++){
19                     if(seg[i][k] && seg[i+k][len-k]){
20                         seg[i][len] = true;
21                         break;
22                     }
23                 }
24             }
25         }
26         return seg[0][length];
27     }

这题貌似是一道面试题：http://thenoisychannel.com/2011/08/08/retiring-a-great-interview-problem/