WordBreak.java

package leetcode;

import java.util.*;

/**
 * @author eko
 *
 * Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
 *
 * Note:
 *
 * The same word in the dictionary may be reused multiple times in the segmentation.
 * You may assume the dictionary does not contain duplicate words.
 * Example 1:
 *
 * Input: s = "leetcode", wordDict = ["leet", "code"]
 * Output: true
 * Explanation: Return true because "leetcode" can be segmented as "leet code".
 * Example 2:
 *
 * Input: s = "applepenapple", wordDict = ["apple", "pen"]
 * Output: true
 * Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
 *              Note that you are allowed to reuse a dictionary word.
 * Example 3:
 *
 * Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
 * Output: false
 *
 */
public class WordBreak {
    public static void main(String[] args) {
        String s = "applepenapple";
        List<String> dic = new ArrayList<>();
        dic.add("apple");
        dic.add("pen");
        System.out.println(new WordBreak().wordBreak(s, dic));
    }

    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> dic = new HashSet<>();
        for (int i = 0; i < wordDict.size(); i++) {
            dic.add(wordDict.get(i));
        }
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && dic.contains(s.substring(j,i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }
}