問題

解法

dp , 狀態dp[i] = true 表示s.substr(i) 是能被分開. 遞推公式為：
dp[i] = dp[i+1] && InDict(s.substr(i, 1)) || dp[i+2] && InDict(s.substr(i, 2)) …
當求出dp[] 陣列後， 可以使用dfs 查詢所有結果。

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
        bool
 
 dp[s.size()+1];
        dp[s.size()] = true;
        for (int i=s.size()-1; i>=0; --i)
        {
            dp[i] = false;
            for (int j = 1; i+j<=s.size(); ++j)
            {
                if (dp[i+j])
                {
                    string now = s.substr(i, j);
                    if
 
 (wordDict.find(now)!= wordDict.end())
                    {
                        dp[i] = true;
                        break;
                    }
                }
            }
        }
        vector<string> ret;
        if (dp[0] == false)
            return ret;
        findResult(ret, 0
 
, "", dp, s, wordDict);
        return ret;
    }
private:
    void findResult(vector<string> &ret, int pos, string curResult, bool dp[], string& s, unordered_set<string>& wordDict)
    {
        if (pos == s.size())
        {
            // elimate last space;
            curResult.pop_back();
            ret.push_back(curResult);
            return;
        }

        for (int i=1; i+pos <=s.size(); ++i)
        {
            if (dp[pos+i])
            {
                string now = s.substr(pos, i);
                if (wordDict.find(now) != wordDict.end())
                {
                    findResult(ret, pos+i, curResult + now + " ", dp, s, wordDict);
                }
            }
        }
    }
};