DP-139-Word Break

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. You may assume the dictionary does not contain duplicate words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.
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class Solution {
    public boolean wordBreak(String str, List<String> wordDict) {
        final int len = str.length();
        boolean[] dp = new boolean[len + 1]; //dp[i]表示到i为止(包括i)的字符串都能构成
        dp[0] = true;
        for(int i = 1; i < dp.length; i++){ //需要填写的dp数组的下标, 能够构成的字符串的长度, i - 1表示到i为止(包括i)的字符串都能构成
            for(int j = 0; j < i; j++){ //字符串起始下标, dp[j]表示起始下标前的字符串能不能构成
                if(dp[j] && wordDict.contains(str.substring(j, i))){
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[len];
    }
}
//dp
//[j]为true
//[j] [j+1] [j+2] ... (i)
//str
//73% O(n2)