【leetcode】第 182 場周賽

可能因爲剛應付完一系列面試，這週末狀態明顯鬆懈，比賽的時候心態也很佛。。比賽過程不多解釋，就着重寫題目分析了。

賽後總結

不足

1. 容器使用還是不夠透徹，map的初始化要搞清楚。

優點

1. 簡單題做的很快。

2. 心態是真的好，不過有點過於佛了。。也不利於思考。

 

題目分析

1. 【easy】1394. Find Lucky Integer in an Array

Given an array of integers arr, a lucky integer is an integer which has a frequency in the array equal to its value.

Return a lucky integer in the array. If there are multiple lucky integers return the largest of them. If there is no lucky integer return -1.

Example 1:

Input: arr = [2,2,3,4]
Output: 2
Explanation: The only lucky number in the array is 2 because frequency[2] == 2.

Example 2:

Input: arr = [1,2,2,3,3,3]
Output: 3
Explanation: 1, 2 and 3 are all lucky numbers, return the largest of them.

Example 3:

Input: arr = [2,2,2,3,3]
Output: -1
Explanation: There are no lucky numbers in the array.

Example 4:

Input: arr = [5]
Output: -1

Example 5:

Input: arr = [7,7,7,7,7,7,7]
Output: 7

Constraints:

• 1 <= arr.length <= 500
• 1 <= arr[i] <= 500

題目鏈接：https://leetcode-cn.com/problems/find-lucky-integer-in-an-array/

思路

根據題意，只需要 數值=出現次數 ，那麼遍歷統計即可。

class Solution {
public:
    int findLucky(vector<int>& arr) {
        int num = arr.size();
        if(num==0) return -1;
        int res = -1;
        map<int,int> rec;
        for(int i=0; i<num; ++i){
            ++rec[arr[i]];
        }
        for(auto iter=rec.rbegin(); iter!=rec.rend(); ++iter){
            if(iter->first == iter->second){
                res = iter->first;
                break;
            }
        }
        return res;
    }
};

 

2.【medium】1395. Count Number of Teams

There are n soldiers standing in a line. Each soldier is assigned a unique rating value.

You have to form a team of 3 soldiers amongst them under the following rules:

• Choose 3 soldiers with index (i, j, k) with rating (rating[i], rating[j], rating[k]).
• A team is valid if:  (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).

Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).

Example 1:

Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1). 

Example 2:

Input: rating = [2,1,3]
Output: 0
Explanation: We can't form any team given the conditions.

Example 3:

Input: rating = [1,2,3,4]
Output: 4

Constraints:

• n == rating.length
• 1 <= n <= 200
• 1 <= rating[i] <= 10^5

題目鏈接：https://leetcode-cn.com/problems/count-number-of-teams/

思路

題目要求選出的3個數字遞增或遞減，但是數列本身不是有序的。

設計一個dp，dp[i]記錄第i位前面小於rating[i]的數字個數，因爲數列並不有序，因此要對i前面所有位置j進行一次判斷。

1）若rating[j]<rating[i]，則++dp[i]，res+=dp[j]，此時可以看作找遞增組，組的數量就是dp[j]的數值；

2）若rating[j]>rating[i]，則res+=(j-dp[j])，此時是找遞減組，組的數量是j位置前大於rating[j]的元素個數，用下標-dp[j]求得。

其實，這個思路代碼有bug，計算遞減組的時候沒有考慮到有相等元素的情況，比如：[2,5,3,5,4,1]。

雖然leetcode上能通過，但最好還是遞減也單獨進行記錄。下面寫的是鑽了leetcode漏洞的版本。

class Solution {
public:
    int numTeams(vector<int>& rating) {
        int num = rating.size();
        if(num<3) return 0;
        int res = 0;
        vector<int> dp(num,0);  // 左邊小於它的數字個數
        for(int i=1; i<num; ++i){
            for(int j=i-1; j>=0; --j){
                if(rating[j]<rating[i]){
                    ++dp[i];
                    res += dp[j];
                }
                if(rating[j]>rating[i]){
                    res += (j-dp[j]);
                }
            }
        }
        return res;
    }
};

 

3.【medium】1396. Design Underground System

Implement the class UndergroundSystem that supports three methods:

1. checkIn(int id, string stationName, int t)

• A customer with id card equal to id, gets in the station stationName at time t.
• A customer can only be checked into one place at a time.

2. checkOut(int id, string stationName, int t)

• A customer with id card equal to id, gets out from the station stationName at time t.

3. getAverageTime(string startStation, string endStation) 

• Returns the average time to travel between the startStation and the endStation.
• The average time is computed from all the previous traveling from startStation to endStation that happened directly.
• Call to getAverageTime is always valid.

You can assume all calls to checkIn and checkOut methods are consistent. That is, if a customer gets in at time t1 at some station, then it gets out at time t2 with t2 > t1. All events happen in chronological order.

Example 1:

Input
["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]

Output
[null,null,null,null,null,null,null,14.0,11.0,null,11.0,null,12.0]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(45, "Leyton", 3);
undergroundSystem.checkIn(32, "Paradise", 8);
undergroundSystem.checkIn(27, "Leyton", 10);
undergroundSystem.checkOut(45, "Waterloo", 15);
undergroundSystem.checkOut(27, "Waterloo", 20);
undergroundSystem.checkOut(32, "Cambridge", 22);
undergroundSystem.getAverageTime("Paradise", "Cambridge");       // return 14.0. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.0
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 12.0

Constraints:

• There will be at most 20000 operations.
• 1 <= id, t <= 10^6
• All strings consist of uppercase, lowercase English letters and digits.
• 1 <= stationName.length <= 10
• Answers within 10^-5 of the actual value will be accepted as correct.

題目鏈接：https://leetcode-cn.com/problems/design-underground-system/

思路

這道題就是根據題目邏輯進行模擬即可，沒有任何坑。

需要一個空間用於記錄進站情況，一個空間記錄平均用時情況。

class UndergroundSystem {
public:
    unordered_map<int, pair<string, int>> in;
    unordered_map<string, unordered_map<string, pair<int, double>>> time;   // num, time
    UndergroundSystem() {

    }
    
    void checkIn(int id, string stationName, int t) {
        in[id] = make_pair(stationName, t);
    }
    
    void checkOut(int id, string stationName, int t) {
        string start = in[id].first;
        double usetime = t*1.0 - in[id].second;
        if(time.find(start)!=time.end() && time[start].find(stationName)!=time[start].end()){
            int num = time[start][stationName].first;
            double oldtime = time[start][stationName].second;
            time[start][stationName].second = (num * oldtime + usetime) / ( num + 1.0);
            ++time[start][stationName].first;
        }else{
            time[start][stationName] = make_pair(1, usetime);
        }
        in.erase(id);
    }
    
    double getAverageTime(string startStation, string endStation) {
        return time[startStation][endStation].second;
    }
};

/**
 * Your UndergroundSystem object will be instantiated and called as such:
 * UndergroundSystem* obj = new UndergroundSystem();
 * obj->checkIn(id,stationName,t);
 * obj->checkOut(id,stationName,t);
 * double param_3 = obj->getAverageTime(startStation,endStation);
 */