2225. Find Players With Zero or One Losses

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題目敘述

You are given an integer array matches where matches[i] = [winneri, loseri] indicates that the player winneri defeated player loseri in a match.

Return a list answer of size 2 where:

answer[0] is a list of all players that have not lost any matches.
answer[1] is a list of all players that have lost exactly one match.

The values in the two lists should be returned in increasing order.

Note:

You should only consider the players that have played at least one match.
The testcases will be generated such that no two matches will have the same outcome.

Example 1

Input: matches = [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]] Output: [[1,2,10],[4,5,7,8]] Explanation: Players 1, 2, and 10 have not lost any matches. Players 4, 5, 7, and 8 each have lost one match. Players 3, 6, and 9 each have lost two matches. Thus, answer[0] = [1,2,10] and answer[1] = [4,5,7,8].

Example 2

Input: matches = [[2,3],[1,3],[5,4],[6,4]] Output: [[1,2,5,6],[]] Explanation: Players 1, 2, 5, and 6 have not lost any matches. Players 3 and 4 each have lost two matches. Thus, answer[0] = [1,2,5,6] and answer[1] = [].

解題思路

Solution

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<List<Integer>> findWinners(int[][] matches) {
        List<List<Integer>> ans = new ArrayList<>();
        List<Integer> noLost = new ArrayList<>();
        List<Integer> oneLost = new ArrayList<>();
        Map<Integer, Integer> map = new HashMap<>();

        for(int[] matche : matches){
            if(!map.containsKey(matche[0])){
                map.put(matche[0], 0);
            }

            if(!map.containsKey(matche[1])){
                map.put(matche[1], 1);
            }else if(map.get(matche[1]) == 0){
                map.put(matche[1], 1);
            }else{
                map.put(matche[1], 2);
            }
        }

        for(int m : map.keySet()){
            if(map.get(m) == 0){
                noLost.add(m);
            }else if(map.get(m) == 1){
                oneLost.add(m);
            }
        }

        Collections.sort(noLost);
        Collections.sort(oneLost);
        ans.add(noLost);
        ans.add(oneLost);

        return ans;
    }
}