【记忆化递归+DP】LeetCode 139. Word Break

LeetCode 139. Word Break

Solution1：
记忆化递归的典型套路题
参考网址：https://zxi.mytechroad.com/blog/leetcode/leetcode-139-word-break/

class Solution { //记忆化递归！
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        return my_word(s, dict);
    }

private:
    unordered_map<string, bool> word_map;

    bool my_word(string s, unordered_set<string>& dict) {
        if (word_map.count(s)) return word_map[s]; //已经判断过的子串
        if (dict.count(s)) 
            return word_map[s] = true;
        for (int i = 1; i < s.size(); i++) {
            string temp1 = s.substr(0, i);
            string temp2 = s.substr(i);
            if (my_word(temp1, dict) && dict.count(temp2)) 
                return word_map[s] = true;
        }
        return word_map[s] = false;
    }    
};

Solution2：
此题也是典型的动态规划题目
参考网址：https://www.youtube.com/watch?v=il8Oi21WZN0
时间复杂度：$O(n^2--n^3)$
空间复杂度：$O(n)$

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int len = s.length();
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        vector<bool> dp(len + 1, false);
        s = " " + s;
        dp[0] = true;
        for (int i = 1; i <= len; i++) {
            for (int j = 0; j < i; j++) {
                string temp = s.substr(j + 1, i - j);
                if (dp[j] && dict.count(temp)) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[len];
    }
};