tiktokOA 1048. Longest String Chain

自己写的，超时

class Solution {
public:
    int longestStrChain(vector<string>& words) {
        int maxn = 1;
        for(int i = 0; i < words.size(); i++){
            maxn = max(maxn, dfs(words[i],words));
        }
        return maxn;
    }
    int dfs(string a, vector<string>& words){
        int maxn = 1;
        for(int i = 0; i < words.size(); i++){
            if(isPre(a, words[i])){
                maxn = max(maxn, 1+dfs(words[i],words));
            }
        }
        return maxn;
    }
    bool isPre(string a, string b){
        if(a.length()+1 != b.length()){
            return false;
        }
        for(int i = 0; i < a.length(); i++){
            if(a[i] != b[i]){
                for(int j = 0; j < 26; j++){
                    char c = 'a' + j;
                    a = a.substr(0,i) + c + a.substr(i);
                    if(a == b)
                        return true;
                    else
                        a = a.substr(0,i) + a.substr(i+1);
                }
                return false;
            }
        }
        return true;
    }
};

剪枝以后依旧超时，但是快了一些

class Solution {
public:
    int longestStrChain(vector<string>& words) {
        unordered_map<int,int> map;
        int maxn = 1;
        for(int i = 0; i < words.size(); i++){
            if(!map[i]){
                map[i] = dfs(i,words,map);
            }
            maxn = max(maxn, map[i]);
        }
        return maxn;
    }
    int dfs(int index, vector<string>& words,unordered_map<int,int> &map){
        int maxn = 1;
        for(int i = 0; i < words.size(); i++){
            if(isPre(words[index], words[i])){
                if(!map[i]){
                    map[i] = dfs(i,words,map);
                }
                maxn = max(maxn, 1+map[i]);
            }
        }
        return maxn;
    }
    bool isPre(string a, string b){
        if(a.length()+1 != b.length()){
            return false;
        }
        for(int i = 0; i < a.length(); i++){
            if(a[i] != b[i]){
                for(int j = 0; j < 26; j++){
                    char c = 'a' + j;
                    a = a.substr(0,i) + c + a.substr(i);
                    if(a == b)
                        return true;
                    else
                        a = a.substr(0,i) + a.substr(i+1);
                }
                return false;
            }
        }
        return true;
    }
};

答案自底向上

class Solution {

public :

    int longestStrChain(vector<string> &words) {
        unordered_map<string, int> dp;

        // Sorting the list in terms of the word length.
        std::sort(words.begin(), words.end(), [](const std::string &word1, const std::string &word2) {
            return word1.size() < word2.size();
        });

        int longestWordSequenceLength = 1;

        for (const string &word : words) {
            int presentLength = 1;
            // Find all possible predecessors for the current word by removing one letter at a time.
            for (int i = 0; i < word.length(); i++) {
                string predecessor = word.substr(0, i) + word.substr(i + 1, word.length() + 1);
                if (dp.find(predecessor) != dp.end()) {
                    int previousLength = dp[predecessor];
                    presentLength = max(presentLength, previousLength + 1);
                }
            }
            dp[word] = presentLength;
            longestWordSequenceLength = max(longestWordSequenceLength, presentLength);
        }
        return longestWordSequenceLength;
    }
};

答案自顶向下

class Solution {
    
private:

    int dfs(unordered_set<string> &words, unordered_map<string, int> &memo, string currentWord) {
        // If the word is encountered previously we just return its value present in the map (memoization).
        if (memo.find(currentWord) != memo.end()) {
            return memo[currentWord];
        }
        // This stores the maximum length of word sequence possible with the 'currentWord' as the
        int maxLength = 1;

        // creating all possible strings taking out one character at a time from the `currentWord`
        for (int i = 0; i < currentWord.length(); i++) {
            string newWord = currentWord.substr(0, i) + currentWord.substr(i + 1);
            // If the new word formed is present in the list, we do a dfs search with this newWord.
            if (words.find(newWord) != words.end()) {
                int currentLength = 1 + dfs(words, memo, newWord);
                maxLength = max(maxLength, currentLength);
            }
        }
        memo[currentWord] = maxLength;

        return maxLength;
    }

public :
    int longestStrChain(vector<string> &words) {
        unordered_map<string, int> memo;
        unordered_set<string> wordsPresent;
        for (const string &word : words) {
            wordsPresent.insert(word);
        }
        int ans = 0;
        for (const string &word : words) {
            ans = max(ans, dfs(wordsPresent, memo, word));
        }
        return ans;
    }
};

答案是改成哈希表了，这样查找快。

修改后自己的代码

class Solution {
public:
    int longestStrChain(vector<string>& words) {
        unordered_map<string,int> map;
        unordered_set<string> wordsPresent;
        for (const string &word : words) {
            wordsPresent.insert(word);
        }
        int maxn = 0;
        for(int i = 0; i < words.size(); i++){
            if(!map[words[i]]){
                map[words[i]] = dfs(words[i],wordsPresent,map);
            }
            maxn = max(maxn, map[words[i]]);
        }
        return maxn;
    }
    int dfs(string s, unordered_set<string>& words,unordered_map<string,int> &map){
        int maxn = 1;
        for(int i = 0; i < s.length(); i++){
            string temp = s.substr(0,i)+s.substr(i+1);
            if(words.find(temp) != words.end()){
                if(!map[temp]){
                    map[temp] = dfs(temp, words, map);
                }
                maxn = max(maxn, 1+map[temp]);
            }
        }
        return maxn;
    }
};

标签：map,return,Chain,int,1048,maxn,Longest,words,word
来源： https://blog.csdn.net/HuiFeiDeTuoNiaoGZ/article/details/120593600