(Trie + DP) 623. K Edit Distance

 

思路：前缀树来存储字符串的所有前缀。

 1. 状态转移方程：

 2. 计算顺序

class Solution {
public:
    /**
     * @param words: a set of stirngs
     * @param target: a target string
     * @param k: An integer
     * @return: output all the strings that meet the requirements
     */
    struct TrieNode{
        TrieNode* child[26];
        bool is_word;
        string str;
        TrieNode(): is_word(false), str(""){
            for(auto& a:child)
                a = NULL;
        }
        ~TrieNode(){
            delete[] child;
        }
    };
    
    struct Trie{
        TrieNode* root = new TrieNode();
        void insert(string word){
            TrieNode* p = root;
            for(char a: word){
                int i = a - 'a';
                if(!p->child[i])
                    p->child[i] = new TrieNode();
                p = p->child[i];
            }
            p->is_word = true;
            p->str = word;
        }
    };
    
    //at node p, prefix Sp
    //f means f[Sp][0~n]
    //todo: Sp -> (Sp,'a')...(Sp,'z') 推导从Sp到更长的前缀
    void dfs(TrieNode* p, vector<int>& f, vector<string>& res, string& target, int k){
        int n = target.size();
        vector<int> nf(n+1, 0);   //新的f
        //whether p has word
        if(p->is_word && f[n] <= k){
            //p是一个完整的单词
            res.push_back(p->str);
        }
        
        //f[Sp][j]:一个前缀Sp和Target前j个字符的最小编辑距离
        //f[i][0] = i
        //f[0] = f[Sp][0] = |Sp|
        nf[0] = f[0] + 1;
 
        for(int i=0;i<26; i++){
            //next char is i
            if(p->child[i] == NULL)
                continue;
            //拓展它的儿子
            for(int j=1; j<=n; j++){
                //f[i][j] = min( f[i][j-1]+1), f[i-1][j]+1, f[i-1][j-1]+1
                nf[j] = min(min(nf[j-1]+1, f[j]+1), f[j-1]+1);
                int c = target[j-1] - 'a';
                if(c == i){
                //f[i][j] = min(f[i][j], f[i-1][j-1]), when A[i-1]==target[j-1]
                   nf[j] = min(nf[j], f[j-1]); 
                }
            }
            dfs(p->child[i], nf, res, target, k);
        }
        
    }
    
    
    vector<string> kDistance(vector<string> &words, string &target, int k) {
        // write your code here
        int n = target.size();
        vector<int> f(n+1,0);
        vector<string> result;
        
        //init Trie
        Trie trie;
        for(string& word: words){
            trie.insert(word);
        }
        TrieNode * p = trie.root;
        //init f
        //f[""][0~n]  前缀为空串时与Target的前n+1个字符的最小编辑距离
        for(int i=0; i<=n; i++){
            f[i] = i;
        }
        
        //dfs
        dfs(p, f, result, target, k);
        //return result
        return result;
    }
};