假设我们有一个数组，其中包含n个元素，从1到n的顺序被打乱。给定另一个整数K。有N个人排队打羽毛球。前两个玩家将去打球，然后失败者将去排队的末尾。胜者将与队列中的下一个人比赛，依此类推。他们将一直打球，直到有人连续赢得K次。然后该选手成为胜者。

如果队列是[2, 1, 3, 4, 5]，K = 2，那么输出将是5。现在看一下解释：

(2, 1)比赛，2获胜，所以1将被添加到队列中，队列变为[3, 4, 5, 1] (2, 3)比赛，3获胜，所以2将被添加到队列中，队列变为[4, 5, 1, 2] (3, 4)比赛，4获胜，所以3将被添加到队列中，队列变为[5, 1, 2, 3] (4, 5)比赛，5获胜，所以4将被添加到队列中，队列变为[1, 2, 3, 4] (5, 1)比赛，5获胜，所以3将被添加到队列中，队列变为[2, 3, 4, 1]

(2, 1)比赛，2获胜，所以1将被添加到队列中，队列变为[3, 4, 5, 1]

(2, 3)比赛，3获胜，所以2将被添加到队列中，队列变为[4, 5, 1, 2]

(3, 4)比赛，4获胜，所以3将被添加到队列中，队列变为[5, 1, 2, 3]

(4, 5)比赛，5获胜，所以4将被添加到队列中，队列变为[1, 2, 3, 4]

(5, 1)比赛，5获胜，所以3将被添加到队列中，队列变为[2, 3, 4, 1]

由于5连续赢得两场比赛，所以输出是5。

算法

winner(arr, n, k)

Begin
   if k >= n-1, then return n
   best_player := 0
   win_count := 0
   for each element e in arr, do
      if e > best_player, then
         best_player := e
         if e is 0th element, then
            win_count := 1
         end if
      else
         increase win_count by 1
      end if
      if win_count >= k, then
         return best player
     done
   return best player
End

Example

示例

#include <iostream>
using namespace std;
int winner(int arr[], int n, int k) {
   if (k >= n - 1) //if K exceeds the array size, then return n
      return n;
   int best_player = 0, win_count = 0; //initially best player and win count is not set
   for (int i = 0; i < n; i++) { //for each member of the array
      if (arr[i] > best_player) { //when arr[i] is better than the best one, update best
         best_player = arr[i];
         if (i) //if i is not the 0th element, set win_count as 1
         win_count = 1;
      }else //otherwise increase win count
      win_count += 1;
      if (win_count >= k) //if the win count is k or more than k, then we have got result
         return best_player;
   }
   return best_player; //otherwise max element will be winner.
}
main() {
   int arr[] = { 3, 1, 2 };
   int n = sizeof(arr) / sizeof(arr[0]);
   int k = 2;
   cout << winner(arr, n, k);
}

输出

3