【2022/04/03-第287场单周赛】复盘

总结

一开始做的时候没看懂第四题。

Q1.转化时间需要的最少操作数

贪心。

class Solution {
public:
    int convertTime(string cu, string co) {
        string ah = {cu[0],cu[1]}, am = {cu[3], cu[4]};
        string bh = {co[0], co[1]}, bm = {co[3], co[4]};
        int x = (stoi(bh) - stoi(ah)) * 60 + stoi(bm) - stoi(am), ret = 0;
        cout << x;
        if(x >= 60){
            ret += x /60;
            x %= 60;
        } 
        if(x >= 15){
            ret += x /15;
            x %= 15;
        }
        if(x >= 5){
            ret += x / 5;
            x %= 5;
        }
        ret += x;
        return ret;
    }
};

Q2.找出输掉零场或一场比赛的玩家

直接哈希。

class Solution {
public:
    vector<vector<int>> findWinners(vector<vector<int>>& matches) {
        int ls[100010] = {0}, w[100010] = {0};
        for(auto i : matches){
            ++ls[i[1]];
            w[i[0]] = 1;
        }
        vector<vector<int>> ret(2);
        for(int i = 0; i < 100010; ++i){
            if(ls[i] == 1) ret[1].push_back(i);
            if(ls[i] == 0 && w[i]) ret[0].push_back(i);
        }
        return ret;
    }
};

Q3.每个小孩最多能分到多少糖果

从0到最大堆二分。

class Solution {
public:
    
    bool canD(vector<int>& candies, long long k, long long mid){
        long long ret = 0;
        for(auto i : candies) ret += i / mid;
        // cout << ret << endl;
        return ret >= k;
    }
    
    int maximumCandies(vector<int>& candies, long long k) {
        long long total = 0;
        // cout << canD(candies, k, 1);
        for(auto i : candies) total += i;
        cout << total;
        if(total < k) return 0;
        long long l = 0, r = total / k;
        while(l < r){
            if(r - l == 1){
                if(canD(candies, k, r)){
                    l = r;
                    break;
                }
                else break;
            }
            // cout << l << ' ' << r << endl; 
            long long mid = (l + r) / 2;
            if(canD(candies, k, mid)) l = mid;
            else r = mid - 1;
        }
        
        
        return l;
    }
};

Q4.加密解密字符串

第三个函数只要把dictionary中所有都加密一次，如果加密后等于需要解密的字符串且长度为待解密字符串长度一半，即算作一个。

class Encrypter {
public:
    
    unordered_map<char, string> cs;
    unordered_map<string, string> ss;
    
    Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary) {
        for(int i = 0; i < keys.size(); ++i) cs[keys[i]] = values[i];
        for(auto s : dictionary) ss[s] = encrypt(s);
    }
    
    string encrypt(string word1) {
        string ret;
        for(auto c : word1) ret += cs[c];
        return ret;
    }
    
    int decrypt(string word2) {
        int ret = 0;
        for(auto i : ss) if(i.second == word2 && i.first.size() == word2.size() / 2) ++ret;
        return ret;
    }
};