Word Break

题目：

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".

这题拿到就开始硬做。。是不对滴。用DP做怎样？搞清楚state，function，initialize和result。

public boolean wordBreak(String s, Set<String> dict) {
        int maxWordLen = 0;
        for (String str : dict) {
            maxWordLen = Math.max(maxWordLen, str.length());
        }
        //F[i] 代表在i时，前面的是否可以被字典中的词拆分
        //F[i] = (F[j] && dict.contains(s.substring(j ,i)))
        //F[0] = true;
        //return F[s.length()]
        boolean[] F = new boolean[s.length() + 1];
        F[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            F[i] = false;
            for (int j = 0; j < i && j <= maxWordLen; j++) {
                if (F[i - j - 1] && dict.contains(s.substring(i - j - 1, i))){
                    F[i] = true;
                }
            }
        }
        return F[s.length()];
    }