leetcode_1048. Longest String Chain_[DP，动态规划，记忆化搜索]

Posted 2021-12-01 jasonlixuetao

https://leetcode.com/problems/longest-string-chain/

Let‘s say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2.  For example, "abc" is a predecessor of "abac".

A word chain is a sequence of words [word_1, word_2, ..., word_k] with k >= 1, where word_1 is a predecessor of word_2, word_2 is a predecessor of word_3, and so on.

Return the longest possible length of a word chain with words chosen from the given list of words.

 

解法：动态规划

对于任意word，任意删去其中一个字母后为word‘，若word‘在words中，则dp[word] = max(dp[word], dp[word‘] + 1)。

先将所有words按长度存储。在计算dp[word]时，先将words中长度为word.size()-1的单词放入一个set，方便word‘是否在words中。

class Solution
{
public:
    int longestStrChain(vector<string>& words)
    {
        vector<vector<string>> len_word(17, vector<string>());
        for(auto word:words)
            len_word[word.size()].push_back(word);
        map<string,int> dp;
        int res=0;
        for(int i=16;i>=1;i--)
        {
            if(i<res)
                break;
            for(auto word:len_word[i])
                res = max(res,dfs(len_word,word,dp));
        }
        return res;
    }
    int dfs(vector<vector<string>>& len_word,string& nowword, map<string,int>& dp)
    {
        //cout<<nowword<<endl;
        if(dp.count(nowword))
            return dp[nowword];
        set<string> Set;
        for(auto word:len_word[nowword.size()-1])
            Set.insert(word);
        int nowres=1;
        for(int i=0; i<nowword.size(); i++)
        {
            string temp=nowword;
            temp.erase(i,1);
            if(Set.count(temp))
                nowres = max(nowres,dfs(len_word,temp,dp)+1);
        }
        return dp[nowword]=nowres;
    }
};