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 playerend
example的中文翻译为：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
以上就是数组元素通过单个移动移动了k个位置？的详细内容。