[leetcode]139. 单词拆分

1、递归+备忘录

class Solution {
    string s;
    vector<string> wordDict;
    unordered_set<string>m_set;
    vector<int>memo;//memo[i]表示[i, size()-1]可不可拆分若干个单词连着 1可 0不可 -1不知道
    int dfs(int start)
    {
        if(start == s.size())
        {
            return 1;
        }
        if(memo[start] != -1)
        {
            return memo[start];
        }
        for(int end = s.size(); end > start; end--)
        {//官方写的是这样for(int end = start; end <= s.size(); end++)
            string temp = s.substr(start,end - start);
            if(m_set.count(temp)==1 && dfs(end) == 1)
            {
                return memo[start] = 1;
            }
        }
        return memo[start] = 0;
    }
public:
    bool wordBreak(string s, vector<string>& wordDict) {
       this->s = s;
       this->wordDict = wordDict;
       this->memo = vector<int>(s.size(), -1); //-1表示没有访问过
       m_set = unordered_set<string>(wordDict.begin(), wordDict.end());
       return dfs(0) == 1;
    }
};

2、BFS：

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string>m_set(wordDict.begin(), wordDict.end());
        queue<int>q;
        vector<bool>visited(s.size(), false);
        q.push(0);
        while(!q.empty())
        {
            int start = q.front();
            q.pop();
            if(visited[start] == false)
            {
                for(int end = s.size(); end > start; end--)
                {//官方写是这样for(int end = start; end <= s.size(); end++)
                    if(m_set.count(s.substr(start, end - start)) == 1)
                    {
                        q.push(end);
                        if(end == s.size())
                        {
                            return true;
                        }
                    }
                }
                visited[start] = true;
            }
        }
        return false;
    }
};

动态规划:

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string>m_set(wordDict.begin(), wordDict.end());
        vector<bool>dp(s.size()+1, false);//dp[i]表示s中以i-1结尾能否被拆分
        dp[0] = true;
        for(int i = 1; i <= s.size(); i++)
        {
            for(int j = 0; j < i; j++)
            {
                if(dp[j] && m_set.count(s.substr(j, i - j)) == 1)
                {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.size()];
    }
};

动态规划优化：

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string>m_set(wordDict.begin(), wordDict.end());
        vector<bool>dp(s.size()+1, false);//dp[i]表示s中[i,size()-1]能否被拆分
        dp[s.size()] = true;
        for(int i = s.size(); i >= 1; i--)
        {
            if(dp[i] == true)
            {
                for(int j = 0; j < i; j++)
                {
                    if(m_set.count(s.substr(j, i - j)) == 1)
                    {
                        dp[j] = true;
                    }
                }
            }
        }
        return dp[0];
    }
};

参考：

• https://leetcode-cn.com/problems/word-break/solution/dan-ci-chai-fen-by-leetcode/