LeetCode 184场周赛

1

class Solution {
public:
    vector<string> stringMatching(vector<string>& words) {
        vector<string> ans;
        int v[105] = {0};
        for (int i = 0; i < words.size(); i++) { //选取第一个单词为待匹配项
            for (int j = 0; j < words.size(); j++) {
                if (j == i || words[i].length() >= words[j].length()) continue; //子串长度大于匹配字符串
                int flag = words[j].find(words[i]);//在j中查找i
                if (flag != words[j].npos && v[i] != 1) {
                    ans.push_back(words[i]);
                    v[i] = 1;
                }
            }
       }
        return ans;
    }
};

2

class Solution {
public:
    vector<int> processQueries(vector<int>& queries, int m) {
        vector<int> ans;
        vector<int> p;
        for (int j = 1; j <= m; j++) p.push_back(j);
        for (int i = 0; i < queries.size(); i++) {
            int tar = queries[i]; //要查找的元素
            for (int j = 0; j < m; j++) {
                if (tar == p[j]) {
                    ans.push_back(j);
                    p.erase(p.begin() + j);
                    p.insert(p.begin(),tar);
                    break;
                }
            }        
        }     
        return ans;
    }
};

3

class Solution {
private:    
    map<string, string> m_pool = {
                                    {"&quot;", """}, {"&apos;", "'"}, {"&amp;", "&"},
                                    {"&gt;", ">"}, {"&lt;", "<"}, {"&frasl;", "/"}
                                 };
    
public:
    string entityParser(string text) {
        string key;
        string res;
        for (auto achar : text) {
            if (achar == '&') {
                if (!key.empty()) {
                    res += key;
                    key.erase();
                }
                key.push_back(achar);
            } else if (achar != ';') {
                key.push_back(achar);
            } else {
                key.push_back(achar);
                if (m_pool.find(key) != m_pool.end()) {
                    // cout << "" << key << ", " << m_pool[key] << ")" << endl; 
                    res += m_pool[key];
                    key.erase();
                } else {
                    res += key;
                    key.erase();
                }
            }
        }
        if (!key.empty()) {
            res += key;
        }
        
        return res;
    }
};

4

class Solution {
public:
    int numOfWays(int n) {
        int Max_num = 1e9 + 7;
        long int k = 6, m = 6,result = 12,temp;//第一列，第二列，结果总数，临时k值存储
        for(int i = 2;i <= n;++i){
            result = ( ((k * 5 ) % Max_num) + ((m * 4 ) % Max_num) )% Max_num;//计算总数
            temp = k;
            k = (((k * 3) % Max_num) + ((m * 2) % Max_num)) % Max_num;//更新k
            m = (((temp * 2) % Max_num) + ((m * 2) % Max_num)) % Max_num;//更新m
        }
        return result;
    }
};