Try   HackMD

2225.Find Players With Zero or One Losses

tags: Medium,Array,Hash Table,Sorting,Counting

2225. Find Players With Zero or One Losses

題目描述

You are given an integer array matches where matches[i] = [winner i, loser i] 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.

範例

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] = [].

Constraints:

  • 1 <= matches.length <=
    105
  • matches[i].length == 2
  • 1 <=
    winneri
    ,
    loseri
    <=
    105
  • winneri
    !=
    loseri
  • All matches[i] are unique.

解答

Javascript

function findWinners(matches) { const NL = []; const oneLoss = []; const lossCount = []; for (const [winner, loser] of matches) { if (!lossCount[winner]) { lossCount[winner] = 0; } if (!lossCount[loser]) { lossCount[loser] = 1; } else { lossCount[loser]++; } } for (let i = 1; i < lossCount.length; i++) { if (lossCount[i] === 0) { NL.push(i); } else if (lossCount[i] === 1) { oneLoss.push(i); } } return [NL, oneLoss]; }

Time complexity :

O(n+k)
這裡k最大為lossCount.length是100000
Marsgoat Nov 28, 2022

C++

class Solution { public: vector<vector<int>> findWinners(vector<vector<int>>& matches) { int counts[100001] = {0}; for (int i = 0; i < matches.size(); i++) { counts[matches[i][0]] |= 1; counts[matches[i][1]] += 2; } vector<vector<int>> ans(2); for (int i = 1; i <= 100000; i++) { int c = counts[i]; if (c == 0) continue; if (c == 1) ans[0].push_back(i); if (c == 2 || c == 3) ans[1].push_back(i); } return ans; } };

Yen-Chi ChenMon, Nov 28

Python

class Solution: def findWinners(self, matches: List[List[int]]) -> List[List[int]]: winners = [winner for winner, _ in matches] losers = [loser for _, loser in matches] ans = [sorted(set(winners) - set(losers))] counter = Counter(losers) ans.append(sorted([loser for loser, times in counter.items() if times == 1])) return ans

Yen-Chi ChenMon, Nov 28

class Solution: def findWinners(self, matches: List[List[int]]) -> List[List[int]]: res = [] winners = [] losers = [] ht = {} for winner, loser in matches: ht[winner] = ht.get(winner, 0) ht[loser] = ht.get(loser, 0) + 1 for key, val in ht.items(): if val == 0: winners.append(key) elif val == 1: losers.append(key) return [sorted(winners), sorted(losers)]

Kobe BryantMon, Nov 28

Reference

回到題目列表