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

分析題目，是給定的一個字串，是否能用已有的詞表完成分詞切割。

用DP的方法來做。

首先定義 dp[i] 表示 s[0, i] 是否能完成分詞（可以完全分為詞表中存在的詞）。故 dp[0] 表示 s[0, 0]，即空串。

1，初始狀態：dp[0] = true

2，遞推式：i由1到len遍歷，將子字串 s[0, i] 分為 s[0,  j] 與 [j, i]，前串是否已經能分詞由 dp[j] 判斷

    bool wordBreak(string s, vector<string>& wordDict) {
        if(wordDict.size()==0) return false;
        int len = s.size();
        int dict_size = wordDict.size();
        // dp[i]表示字串 s[0,i] (即0~i-1位) 是否能完成分詞。
        // 故dp[0]表示空串，dp長度應為len+1，因為 s[0,len](0~len-1位)應為dp[len+1]來判斷
        vector<bool> dp(len+1, false);
        dp[0] = true;
        // 迴圈遍歷，i從 1到len，已知i處即為字串的i-1處
        for(int i=1; i<=len ; i++) {
            // j 從 i-1到0，即j將子字串[0, i] 分為[0, j] 與 [j, i]
            for(int j=i-1; j>=0; j--) {
                if (dp[j]) {  // 前串[0, j]可分
                    string subword = s.substr(j, i-j);
                    // 若後段存在於字典中
                    if(find(wordDict.begin(), wordDict.end(), subword) != wordDict.end()){  
                        dp[i] = true;
                        break;
                    }
                }
            }
        }
        return dp[len];
    }