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

題解：跟上一篇一樣，難度更簡單，直接遍歷s求出dp陣列就行，dp記錄每個位置之前的字串是否可分，以後bug free要注意呼叫函式引數是否一致，還有是否缺少引用等細節

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        set<string> dict;
        for(auto ss:wordDict){
            dict.insert(ss);
        }
        if(dict.size()==0) return false;
        vector<bool> dp(s.size()+1,false);
        for(int i=0;i<s.size();i++){
            
            for(int j=0;j<=i;j++){
                string str=s.substr(j,i-j+1);
                if(dict.count(str)!=0){
                    if(j==0) dp[i]=true;
                    else{
                        if(dp[j-1]) dp[i]=true;
                    }
                }
            }
        }
        return dp[s.size()-1];
    }
    
};