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] = [].
Constraints:
1 <= matches.length <= 105
matches[i].length == 2
1 <= winneri, loseri <= 105
winneri != loseri
- All
matches[i]
are unique.
Solution: Hashtable
Use a hashtable to store the number of matches each player has lost. Note, also create an entry for those winners who never lose.
Time complexity: O(m), m = # of matches
Space complexity: O(n), n = # of players
C++
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// Author: Huahua class Solution { public: vector<vector<int>> findWinners(vector<vector<int>>& matches) { unordered_map<int, int> m; for (const auto& match : matches) { m[match[0]]; // create an entry for the winner. ++m[match[1]]; } vector<vector<int>> ans(2); for (const auto& [player, loses] : m) if (loses <= 1) ans[loses].push_back(player); sort(begin(ans[0]), end(ans[0])); sort(begin(ans[1]), end(ans[1])); return ans; } }; |
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