标签：problem pos ems res public rds 单词 can chain
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; } };
leetcode_1048. Longest String Chain_[DP，动态规划，记忆化搜索]
标签：problem pos ems res public rds 单词 can chain
原文地址：https://www.cnblogs.com/jasonlixuetao/p/10888985.html
