Leetcode 139. Word Break

遞歸，超時

class Solution {
public:
    using USET = unordered_set<string>;
    bool wordBreak(string s, vector<string>& wordDict) {
        USET st;
        int minLen = INT_MAX;
        int maxLen = INT_MIN;
        for (auto& word : wordDict) {
            int len = word.size();
            if (len > maxLen)
                maxLen = len;
            if (len < minLen)
                minLen = len;
            st.insert(word);            
        }

        return reSolver(s, st, minLen, maxLen);
    }
private:
    bool reSolver(string s, USET& st, int minLen, int maxLen) {
        int len = s.size();
        if (len == 0) return true;
        if (len < minLen) return false;

        int i = len;
        if (i > maxLen)
            i = maxLen;


        for (; i >= minLen; --i) {
            string substr = s.substr(0, i);
            USET::iterator it = st.find(substr);
            if (it == cend(st)) continue;

            if (reSolver(s.substr(i, len - i), st, minLen, maxLen) == true)
                return true;
        }
        return false;
    }

};

動態規劃

class Solution {
public:
    using USET = unordered_set<string>;
    bool wordBreak(string s, vector<string>& wordDict) {
        if (wordDict.size() == 0) return false;
        USET st;
        set<int>lens;
        vector<int> dp(s.size() + 1, 0);

        for (auto& word : wordDict) {
            int len = word.size();

            // 有序的
            lens.insert(len);
            st.insert(word);            
        }
        dp[0] = 1;

        for (int i = 1; i <= s.size(); ++i) {
            for (auto& len : lens) {
                if (i - len < 0) break; 
                string substr = s.substr(i - len, len);
                USET::iterator it = st.find(substr);
                if (it == cend(st)) continue;
                if (dp[i - len] == 1) {
                    dp[i] = 1;
                    break;
                }
            }
        }
        return dp[s.size()];
    }


};