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

  dp[i]表示长度为s的string能不能被segmented，size总共有1~n，再加上size=0，所以dp长度得是n+1，最后返回dp[n]。 对于s[0, i]，开一个小loop从i往前，如果其中dp[i-j]满足条件，当前s[i-j, i]也能在dict里找到，那dp[i]整体就满足条件，只要这段有满足条件的了就可以break出去继续拓展i了。   实现: Time O(n^3)-substrO(n) Space (n)

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        
        unordered_set dict(wordDict.begin(), wordDict.end());
        int n = s.size();
        vector<bool> dp(n+1, false);
        dp[0] = true;
        
        for (int i=1; i<=n; i++){
            for (int j=i; j>0; j--){
                string sub = s.substr(i-j, j);
                if (dict.count(sub) > 0){
                    if (dp[i-j]){
                        dp[i] = true;
                        break;
                    }
                }
            }
        }
        
        return dp[n];
    }
};