There are n friends that are playing a game. The friends are sitting in a circle and are numbered from 1 to n in clockwise order. More formally, moving clockwise from the ith friend brings you to the (i+1)th friend for 1 <= i < n, and moving clockwise from the nth friend brings you to the 1st friend.
The rules of the game are as follows:
1st friend receives the ball.
- After that,
1stfriend passes it to the friend who isksteps away from them in the clockwise direction. - After that, the friend who receives the ball should pass it to the friend who is
2 * ksteps away from them in the clockwise direction. - After that, the friend who receives the ball should pass it to the friend who is
3 * ksteps away from them in the clockwise direction, and so on and so forth.
In other words, on the ith turn, the friend holding the ball should pass it to the friend who is i * k steps away from them in the clockwise direction.
The game is finished when some friend receives the ball for the second time.
The losers of the game are friends who did not receive the ball in the entire game.
Given the number of friends, n, and an integer k, return the array answer, which contains the losers of the game in the ascending order.
Example 1:
Input: n = 5, k = 2 Output: [4,5] Explanation: The game goes as follows: 1) Start at 1st friend and pass the ball to the friend who is 2 steps away from them - 3rd friend. 2) 3rd friend passes the ball to the friend who is 4 steps away from them - 2nd friend. 3) 2nd friend passes the ball to the friend who is 6 steps away from them - 3rd friend. 4) The game ends as 3rd friend receives the ball for the second time.
Example 2:
Input: n = 4, k = 4 Output: [2,3,4] Explanation: The game goes as follows: 1) Start at the 1st friend and pass the ball to the friend who is 4 steps away from them - 1st friend. 2) The game ends as 1st friend receives the ball for the second time.
Constraints:
1 <= k <= n <= 50
找出转圈游戏输家。
n 个朋友在玩游戏。这些朋友坐成一个圈，按 顺时针方向 从 1 到 n 编号。从第 i 个朋友的位置开始顺时针移动 1 步会到达第 (i + 1) 个朋友的位置（1 <= i < n），而从第 n 个朋友的位置开始顺时针移动 1 步会回到第 1 个朋友的位置。
游戏规则如下：
第 1 个朋友接球。
- 接着，第
1个朋友将球传给距离他顺时针方向k步的朋友。 - 然后，接球的朋友应该把球传给距离他顺时针方向
2 * k步的朋友。 - 接着，接球的朋友应该把球传给距离他顺时针方向
3 * k步的朋友，以此类推。
换句话说，在第 i 轮中持有球的那位朋友需要将球传递给距离他顺时针方向 i * k 步的朋友。
当某个朋友第 2 次接到球时，游戏结束。
在整场游戏中没有接到过球的朋友是 输家 。
给你参与游戏的朋友数量 n 和一个整数 k ，请按升序排列返回包含所有输家编号的数组 answer 作为答案。
思路是模拟。这里我们需要创建一个长度为 n 的 boolean 数组记录每个下标是否都被访问过。注意最后让我们输出的不是未访问过的下标，而是下标 + 1。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public int[] circularGameLosers(int n, int k) { 3 boolean[] players = new boolean[n]; 4 int round = 1; 5 int index = 0; 6 while (players[index] == false) { 7 players[index] = true; 8 index = (index + round * k) % n; 9 round++; 10 } 11 12 List<Integer> list = new ArrayList<>(); 13 for (int i = 0; i < n; i++) { 14 if (players[i] == false) { 15 list.add(i); 16 } 17 } 18 int[] res = new int[list.size()]; 19 for (int i = 0; i < list.size(); i++) { 20 res[i] = list.get(i) + 1; 21 } 22 return res; 23 } 24 }