[LeetCode] 2073. Time Needed to Buy Tickets

There are n people in a line queuing to buy tickets, where the 0th person is at the front of the line and the (n - 1)th person is at the back of the line.

You are given a 0-indexed integer array tickets of length n where the number of tickets that the ith person would like to buy is tickets[i].

Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leave the line.

Return the time taken for the person at position k (0-indexed) to finish buying tickets.

Example 1:
Input: tickets = [2,3,2], k = 2
Output: 6
Explanation:

• In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1].
• In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0].
  The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.

Example 2:
Input: tickets = [5,1,1,1], k = 0
Output: 8
Explanation:

• In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0].
• In the next 4 passes, only the person in position 0 is buying tickets.
  The person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.

Constraints:
n == tickets.length
1 <= n <= 100
1 <= tickets[i] <= 100
0 <= k < n

买票需要的时间。

有 n 个人前来排队买票，其中第 0 人站在队伍 最前方 ，第 (n - 1) 人站在队伍 最后方 。

给你一个下标从 0 开始的整数数组 tickets ，数组长度为 n ，其中第 i 人想要购买的票数为 tickets[i] 。

每个人买票都需要用掉 恰好 1 秒 。一个人 一次只能买一张票 ，如果需要购买更多票，他必须走到 队尾 重新排队（瞬间 发生，不计时间）。如果一个人没有剩下需要买的票，那他将会 离开 队伍。

返回位于位置 k（下标从 0 开始）的人完成买票需要的时间（以秒为单位）。

思路

这道题的暴力解是用一个队列，把所有人需要购买的票数放进去，然后开始遍历每一个元素。但是如果人数太多（n 太大）就一定会超时。

这里我提供一个次优解，不需要用队列，就是简单的模拟。当第 k 个人还未完成购票的时候，就一直遍历整个 input 数组。因为每个人一次只能购买一张票，一次购买的动作耗时一秒钟，所以我们一直遍历整个数组直到第 k 个人完成购票，才把循环 break，返回结果。

这里同时我提供一个网上看到的最优解，时间复杂度为O(n)，但是很难想到。

复杂度

时间O(n * k)
空间O(1)

代码

Java实现

class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
		int n = tickets.length;
		int time = 0;
		while (tickets[k] > 0) {
			for (int i = 0; i < n; i++) {
				if (tickets[i] > 0) {
					tickets[i]--;
					time++;
					if (i == k && tickets[k] == 0) {
						return time;
					}
				}
			}
		}
		return time;
    }
}