Lecture 287 of the Week: Players who have lost zero or one game

Get into the habit of writing together! This is the 5th day of my participation in the "Nuggets Daily New Plan · April Update Challenge", click to view the details of the event .

1. Problem description

Gives you an array of integers  matches which  represent  beats  matches[i] = [winneri, loseri] in a game   .winneriloseri

Returns a list of length 2** answer :

• answer[0] is a list of all  players who have not  lost any games.
• answer[1] is a list of all players who happened to lose  a  game.

The values ​​in both lists should be  returned in increasing  order.

Notice:

• Only those players who participated in  at least one  match were considered.
• The generated test case guarantees that  no  two matches  will have the same result  .

Topic link: Players who have lost zero or one game .

Second, the subject requirements

Sample

输入： matches = [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]]
输出： [[1,2,10],[4,5,7,8]]
解释：
玩家 1、2 和 10 都没有输掉任何比赛。
玩家 4、5、7 和 8 每个都输掉一场比赛。
玩家 3、6 和 9 每个都输掉两场比赛。
因此，answer[0] = [1,2,10] 和 answer[1] = [4,5,7,8] 。
复制代码

visit

1.哈希、判重
2.建议用时15~35min
复制代码

3. Problem Analysis

This question is actually very easy to understand. At first, my idea was to use a hash table to count the losers.

If you lose 0 games , you can only judge whether the player who won the game is in it or not. If you lose a game, you can directly judge the value of the hash table to 1 .

However, this method has a flaw. The execution time is too long, and the winner needs to be judged to prevent repeated joining.

Start optimizing!

The map can be sorted automatically, so this time use the map output storage. We first store the winning player in the map, and set the initial value to 0. Then store the losing player to the map, at this time we only need to judge 0 1.

Fourth, the encoding implementation

1. Before optimization

class Solution {
public:
    vector<vector<int>> findWinners(vector<vector<int>>& matches) {
        map<int,int>l,k;//map存储输掉比赛的人数
        int i,j,m=matches.size();//初始化数据
        vector<int>ans1,ans2;//分别存储
        for(i=0;i<m;i++)
            l[matches[i][1]]++;//记录输掉比赛的人数
        for(i=0;i<m;i++)
        {
            if(l[matches[i][0]]==0&&k[matches[i][0]]==0)//没有输
            {
                ans1.push_back(matches[i][0]);
                k[matches[i][0]]=1;//用来判重
            }
            if(l[matches[i][1]]==1&&k[matches[i][1]]==0)//只输一场
            {
                   ans2.push_back(matches[i][1]);
                   k[matches[i][1]]=1;//用来判重
            }
        }  
        sort(ans1.begin(),ans1.end());//排序
        sort(ans2.begin(),ans2.end());//排序
        return {ans1,ans2};//输出结果
    }
};
复制代码

2. After optimization

class Solution {
public:
    vector<vector<int>> findWinners(vector<vector<int>>& matches) {
        map<int,int>m;//定义map
        vector<int>ans1,ans2;
        int i,n=matches.size();//初始数据
        for(i=0;i<n;i++)//存储赢得人数、置为0
            m[matches[i][0]]=0;
        for(i=0;i<n;i++)//存储输人数++
            m[matches[i][1]]++;
        for(auto j=m.begin();j!=m.end();j++)//在map循环内部存储输出
        {
            if(j->second==0)//为0
                ans1.push_back(j->first);//存储没输的玩家
            else if(j->second==1)//为1
                ans2.push_back(j->first);//存储只输一场的玩家
        }
        return {ans1,ans2};
    }
};

复制代码

5. Test results

The code is a lot simpler, but the execution time is still very long. Let’s optimize it next time.