Description

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

分析

题目的意思是：给你一个数组，第0个位置和第1个位置的数比较，较大的数放在第0个位置，较小的数放在最后，其他的数向前移动一个元素。这道题最好的解法是找规律，对与当前的元素cur，只需要在后续遍历的时候找到比他小的k个数就是符合题目条件的数了。所以遍历的时候利用cnt来统计比当前cur小的数目，初始化为0。注意当碰到比当前值大的时候，cur要重新赋值，这时候，cur应初始化为1，应该新的cur在当前已经有了一个比他小的值了哈哈哈哈。还是很巧妙的。

  • 另一种思路就是利用队列来模拟一下该过程，deque，哈哈，有兴趣的可以琢磨一下。利用队列的代码链接给在下面了。

代码

class Solution:
    def getWinner(self, arr: List[int], k: int) -> int:
        n=len(arr)
        if(k>=n):
            return max(arr)
        for i in range(n):
            if(i==0):
                cur=arr[i]
                cnt=0
            elif(cur>=arr[i]):
                cnt+=1
            else:
                cur=arr[i]
                cnt=1
            if(cnt==k):
                break
        return cur

参考文献

​​[LeetCode] Python; O(1) space; O(n) time - Single Pass​​​​[LeetCode]O(N) space time approach using deque in python. simple intuitive approach​​