【leetcode】1535. Find the Winner of an Array Game

题目如下：

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

解题思路：题目不难，首先把arr[0]记为最大值，然后依次与arr[i]进行比较；如果最大值小于arr[i]，则把arr[i]记为新的最大值，在比较的过程中同时记录每次出现的最大值赢的次数即可。如果arr遍历完成，还是没有哪个数字赢得超过k次以上，那么返回arr的最大值，因为最大值一定是可以一直赢下去的。

代码如下:

class Solution(object):
    def getWinner(self, arr, k):
        """
        :type arr: List[int]
        :type k: int
        :rtype: int
        """
        top = arr[0]
        top_win = 0
        for i in range(1,len(arr)):
            if arr[i] > top:
                top = arr[i]
                top_win = 1
            else:
                top_win += 1
            if top_win == k:
                break
        return top