leetcode — word-break

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

/**
 * Source : https://oj.leetcode.com/problems/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".
 *
 */
public class WordBreak {

    /**
     * 判断给定的字符串是否能分割为多个单词，每个单词都包含在给定的字典中
     *
     * 1. 使用DFS，如果能到达字符串最后，说明可以break
     * 2. 题目中只是判断是否的问题，并不需要求出具体break的方法，对于是否的问题可以使用DP来解决
     *
     * dp[i]:表示str[i-1]能否被break，比如dp[1]表示str[0:0]能否被break
     * dp[0] = true
     * dp[i] = true,当：
     * 存在0 <= k <= i-1, dp[k] = true, 并且tr[k:i-1] 存在dic中
     *
     * @param str
     * @param dic
     * @return
     */
    public boolean wordBreak (String str, Set<String> dic) {
        boolean[] dp = new boolean[str.length()+1];
        dp[0] = true;
        for (int i = 0; i < str.length(); i++) {
            for (int j = i; j > -1; j--) {
                if (dp[j] && dic.contains(str.substring(j, i+1))) {
                    dp[i + 1] = true;
                    break;
                }
            }
        }
        return dp[str.length()];
    }


    public static void main(String[] args) {
        WordBreak wordBreak = new WordBreak();
        String[] dicStr = new String[]{"leet", "code"};
        boolean result = wordBreak.wordBreak("leetcode", new HashSet<String>(Arrays.asList(dicStr)));
        System.out.println(result + " == true");
    }
}