139 Word Break *****representative DP time O(n^2) space O(n)

should know more solutions

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

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
       boolean[] dp = new boolean[s.length() + 1];//
       //why length() + 1因为当取SUBSTRING(J,I)i=1取出来第一个字母
        dp[0] = true;

        for (int i = 1; i <= s.length(); i++) {//因为取substring时候，得取到最后一个字母，所以 <=s.length（）
            for (int j = 0; j < i; j++) {
                if(dp[j] && wordDict.contains(s.substring(j,i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }
}

store dp array to true,

set on dp[I] = true