Table of contents
Given problem
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:
 Start at the
1st
friend.  Count the next
k
friends in the clockwise direction including the friend you started at. The counting wraps around the circle and may count some friends more than once.  The last friend you counted leaves the circle and loses the game.
 If there is still more than one friend in the circle, go back to step
2
starting from the friendimmediately clockwise
of the friend who just lost and repeat.  Else, the last friend in the circle wins the game.
Given the number of friends, n
, and an integer k
, return the winner of the game.
Example 1:
Input: n = 5, k = 2
Output: 3
Explanation: Here are the steps of the game:
1) Start at friend 1.
2) Count 2 friends clockwise, which are friends 1 and 2.
3) Friend 2 leaves the circle. Next start is friend 3.
4) Count 2 friends clockwise, which are friends 3 and 4.
5) Friend 4 leaves the circle. Next start is friend 5.
6) Count 2 friends clockwise, which are friends 5 and 1.
7) Friend 1 leaves the circle. Next start is friend 3.
8) Count 2 friends clockwise, which are friends 3 and 5.
9) Friend 5 leaves the circle. Only friend 3 is left, so they are the winner.
Example 2:
Input: n = 6, k = 5
Output: 1
Explanation: The friends leave in this order: 5, 4, 6, 2, 3. The winner is friend 1.
Constraints:
 1 <= k <= n <= 500
Using Circular Linked List
The first approach is to use brute force algorithm. It means that we will create a circular linked list to simulate all steps above.
Below is the source code of this way.
class Solution {
public int findTheWinner(int n, int k) {
CircularLinkedList circularGame = new CircularLinkedList();
for (int i = 1; i <= n; ++i) {
circularGame.insert(i);
}
int count = k;
CircularLinkedList.Node currentNode = circularGame.head();
while (circularGame.hasNotRemainedOne()) {
CircularLinkedList.Node kNode = circularGame.incrementWithSteps(currentNode, count);
currentNode = kNode.next;
circularGame.remove(kNode);
}
return circularGame.head().value;
}
class CircularLinkedList {
private Node head;
private Node tail;
public void insert(int value) {
Node newNode = new Node(value);
if (this.head == null) {
head = newNode;
} else {
this.tail.next = newNode;
}
this.tail = newNode;
this.tail.next = head;
}
public void remove(Node node) {
if (node == null) {
return;
}
Node currentNode = this.head;
do {
if (currentNode.next == node) {
currentNode.next = node.next;
if (this.head == node) {
this.head = this.head.next;
}
if (this.tail == node) {
this.tail = currentNode;
}
break;
}
currentNode = currentNode.next;
} while (currentNode != this.head);
}
public boolean hasNotRemainedOne() {
return this.head != this.tail;
}
public Node incrementWithSteps(Node startNode, int k) {
if (k == 1) {
return startNode;
}
if (k < 1  startNode == null) {
return null;
}
Node currentNode = startNode;
do {
currentNode = currentNode.next;
k;
} while (k != 1);
return currentNode;
}
public Node head() {
return this.head;
}
static class Node {
public Node next;
public int value;
public Node(int value) {
this.value = value;
this.next = null;
}
}
}
}
The complexity of this solution:
 Time complexity: O(n)
 Space complexity: O(n)
Using Queue
To reduce the number of line of code in the above solution, we can use Queue data structure. The logic is still remained.
class Solution {
public int findTheWinner(int n, int k) {
Queue<Integer> circularGame = new LinkedList<>();
for (int i = 1; i <= n; ++i) {
circularGame.add(i);
}
while (circularGame.size() > 1) {
int count = k;
while (count > 1) {
int current = circularGame.poll();
circularGame.add(current);
count;
}
circularGame.poll();
}
return circularGame.peek();
}
}
Wrapping up

When running the two solutions in the leetcode, we found that:
 The way that use CircularLinkedList is running faster than Using Queue data structure, because using Queue took 125 ms, but Using Circular Linked List took 8 ms.
Reference: