139 单词的组成

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        auto wordDictSet = unordered_set <string> (wordDict.begin(). wordDict.end());

        auto dp = vector <bool> (s.size() + 1);
        dp[0] = true;
        for (int i = 1; i <= s.size(); ++i) {
            for (int j = 0; j < i; ++j) {
                if (dp[j] && wordDictSet.find(s.substr(j, i - j)) != wordDictSet.end()) {
                    dp[i] = true;
                    break;
                }
            }
        }

        return dp[s.size()];
    }
};

 

个人错误思路： 递归，理论上是可行的。 不过时间复杂度太高，递归深度太深。

void checkComposeHelp(string s, vector<string> & wordDict, bool & canComposed){
        if(s.length() == 0){
            canComposed = true;
            return;
        }
        for(int i = 0; i < wordDict.size(); ++i){
            int findPos = s.find(wordDict[i]);
            if(findPos == 0 || (findPos != string::npos && findPos + wordDict[i].length() == s.length())){
                string sBak(s);
                sBak = sBak.erase(findPos, wordDict[i].length());
                checkComposeHelp(sBak, wordDict, canComposed);
1
                if (canComposed){
                    return;
                }
            }else{
                continue;
            }
        }
        return;
    }
    bool wordBreak(string s, vector<string>& wordDict) {
        bool canComposed = false;
        sort(wordDict.begin(), wordDict.end(), [](string one, string another){return one.length() > another.length();});
        checkComposeHelp(s,wordDict, canComposed);
        return canComposed;
    }