[数据结构与算法]Leetcode 1823.找出游戏的获胜者

原题链接：Leetcode 1823. 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:

1. Start at the 1st friend.
2. 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.
3. The last friend you counted leaves the circle and loses the game.
4. If there is still more than one friend in the circle, go back to step 2 starting from the friend immediately clockwise of the friend who just lost and repeat.
5. 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

方法一：队列+模拟

思路：

模拟游戏的过程：
首先将玩家加入到一个队列当中，对于给定的K，执行K-1次将队头移动到队尾的操作，此时就是要淘汰的人。将他出队。最后队列中剩下的唯一元素就是胜者，输出即可。

c++代码：

class Solution {
public:
    int findTheWinner(int n, int k) {
        queue<int> Q;
        for(int i = 1; i <= n; i++ ){
            Q.push(i);
        }
        
        while(Q.size() > 1){
            // k-1轮 将元素从队头放到队尾
            for(int i = 1; i < k; i++ ){
                Q.push(Q.front());
                Q.pop();
            }
            // 队头出队 表示淘汰
            Q.pop();
        }
        return Q.front();
    }
};

复杂度分析：

• 时间复杂度：O(nk)，需要淘汰n-1个人，淘汰每个人需要循环k-1次
• 空间复杂度：O(n)，用一个队列来存储元素

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