LintCode-Word Segmentation

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.

Example

Given

s = "lintcode",

dict = ["lint", "code"].

Return true because "lintcode" can be segmented as "lint code".

Analysis:

It is a DP problem. However, we need to use charAt() instead of substring() to optimize speed. Also, we can first check whether each char in s has appeared in dict, if not, then directly return false. (This is used to pass the last test case in LintCode).

Solution:

 public class Solution {
     /**
      * @param s: A string s
      * @param dict: A dictionary of words dict
      */
     public boolean wordSegmentation(String s, Set<String> dict) {
         if (s.length()==0) return true;

         char[] chars = new char[256];
         for (String word : dict)
             for (int i=0;i<word.length();i++)
                 chars[word.charAt(i)]++;

         for (int i = 0;i<s.length();i++)
             if (chars[s.charAt(i)]==0) return false;

         boolean[] d = new boolean[s.length()+1];
         Arrays.fill(d,false);
         d[0] = true;
         for (int i=1;i<=s.length();i++){
         StringBuilder builder = new StringBuilder();
             for (int j=i-1;j>=0;j--){
                 builder.insert(0,s.charAt(j));
                 String cur = builder.toString();
                 if (d[j] && dict.contains(cur)){
                     d[i]=true;
                     break;
                 }
             }
         }

         return d[s.length()];

     }
 }