LeetCode contest 200 5476. 找出数组游戏的赢家 Find the Winner of an Array Game
Table of Contents
一、中文版
给你一个由 不同 整数组成的整数数组 arr 和一个整数 k 。
每回合游戏都在数组的前两个元素(即 arr[0] 和 arr[1] )之间进行。比较 arr[0] 与 arr[1] 的大小,较大的整数将会取得这一回合的胜利并保留在位置 0 ,较小的整数移至数组的末尾。当一个整数赢得 k 个连续回合时,游戏结束,该整数就是比赛的 赢家 。
返回赢得比赛的整数。
题目数据 保证 游戏存在赢家。
示例 1:
输入:arr = [2,1,3,5,4,6,7], k = 2
输出:5
解释:一起看一下本场游戏每回合的情况:
因此将进行 4 回合比赛,其中 5 是赢家,因为它连胜 2 回合。
示例 2:
输入:arr = [3,2,1], k = 10
输出:3
解释:3 将会在前 10 个回合中连续获胜。
示例 3:
输入:arr = [1,9,8,2,3,7,6,4,5], k = 7
输出:9
示例 4:
输入:arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
输出:99
提示:
2 <= arr.length <= 10^5
1 <= arr[i] <= 10^6
arr 所含的整数 各不相同 。
1 <= k <= 10^9
二、英文版
Given an integer array arr of distinct integers and an integer k.
A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0 and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds.
Return the integer which will win the game.
It is guaranteed that there will be a winner of the game.
Example 1: Input: arr = [2,1,3,5,4,6,7], k = 2 Output: 5 Explanation: Let's see the rounds of the game: Round | arr | winner | win_count 1 | [2,1,3,5,4,6,7] | 2 | 1 2 | [2,3,5,4,6,7,1] | 3 | 1 3 | [3,5,4,6,7,1,2] | 5 | 1 4 | [5,4,6,7,1,2,3] | 5 | 2 So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games. Example 2: Input: arr = [3,2,1], k = 10 Output: 3 Explanation: 3 will win the first 10 rounds consecutively. Example 3: Input: arr = [1,9,8,2,3,7,6,4,5], k = 7 Output: 9 Example 4: Input: arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000 Output: 99 Constraints: 2 <= arr.length <= 10^5 1 <= arr[i] <= 10^6 arr contains distinct integers. 1 <= k <= 10^9
三、My answer
class Solution: def getWinner(self, arr: List[int], k: int) -> int: res = 0 n = len(arr) cnt = 0 # 记录每个数胜利的次数 i = 0 idx = 1 while i + idx < n and cnt < min(n,k): if arr[i] < arr[i+idx]: i = i+idx idx = 1 cnt = 1 else: cnt += 1 if cnt == k: return arr[i] idx += 1 return arr[i]
四、解题报告
数据结构:数组
算法:遍历
实现:
1、根据题目限制,不能暴力遍历,肯定会超时,所以想办法只经过一次遍历即可。
2、如果 k 大于 arr 的长度,由于将 arr 中所有数字两两比较之后一定会选出 “赢家”,所以此时不需要遍历 k 次,一个数字只需要赢 min(n,k) 次即可,其中 n 是 arr 的长度。
3、每次比较两个数,如果是连胜,则 count += 1,判断 count 与 k 的关系,如果等于 k,则返回。
4、如果是新的数字,则 idx 和 count 都需要重新归1。 idx 表示当前 i 与往后第 idx 个数比较大小。
如果 i+idx 的数是赢家, i 直接跳到 i + idx 即可,即 i=i+idx。因为期间的数字都败给第 i 个数了