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LeetCode: 1244. 力扣排行榜

1、题目描述

新一轮的「力扣杯」编程大赛即将启动，为了动态显示参赛者的得分数据，需要设计一个排行榜 Leaderboard。

请你帮忙来设计这个 Leaderboard 类，使得它有如下 3 个函数：

  1. addScore(playerId, score)：

    • 假如参赛者已经在排行榜上，就给他的当前得分增加 score 点分值并更新排行。

    • 假如该参赛者不在排行榜上，就把他添加到榜单上，并且将分数设置为 score。

  2. top(K)：返回前 K 名参赛者的 得分总和。

  3. reset(playerId)：将指定参赛者的成绩清零。题目保证在调用此函数前，该参赛者已有成绩，并且在榜单上。 请注意，在初始状态下，排行榜是空的。

示例 1：

输入： 
["Leaderboard","addScore","addScore","addScore","addScore","addScore","top","reset","reset","addScore","top"]
[[],[1,73],[2,56],[3,39],[4,51],[5,4],[1],[1],[2],[2,51],[3]]
输出：
[null,null,null,null,null,null,73,null,null,null,141]

解释： 
Leaderboard leaderboard = new Leaderboard ();
leaderboard.addScore(1,73);   // leaderboard = [[1,73]];
leaderboard.addScore(2,56);   // leaderboard = [[1,73],[2,56]];
leaderboard.addScore(3,39);   // leaderboard = [[1,73],[2,56],[3,39]];
leaderboard.addScore(4,51);   // leaderboard = [[1,73],[2,56],[3,39],[4,51]];
leaderboard.addScore(5,4);    // leaderboard = [[1,73],[2,56],[3,39],[4,51],[5,4]];
leaderboard.top(1);           // returns 73;
leaderboard.reset(1);         // leaderboard = [[2,56],[3,39],[4,51],[5,4]];
leaderboard.reset(2);         // leaderboard = [[3,39],[4,51],[5,4]];
leaderboard.addScore(2,51);   // leaderboard = [[2,51],[3,39],[4,51],[5,4]];
leaderboard.top(3);           // returns 141 = 51 + 51 + 39;

提示：

  • 1 <= playerId, K <= 10000
  • 题目保证 K 小于或等于当前参赛者的数量
  • 1 <= score <= 100
  • 最多进行 1000 次函数调用

2、解题思路

  • 使用collections.Counter实现
from collections import Counter


class Leaderboard:

    def __init__(self):
        self.count = Counter()

    def addScore(self, playerId: int, score: int) -> None:
        self.count[playerId] += score

    def top(self, K: int) -> int:
        return sum(x for i, x in self.count.most_common(K))

    def reset(self, playerId: int) -> None:
        self.count[playerId] = 0

# Your Leaderboard object will be instantiated and called as such:
# obj = Leaderboard()
# obj.addScore(playerId,score)
# param_2 = obj.top(K)
# obj.reset(playerId)