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
買票需要的時間。
有 n 個人前來排隊買票，其中第 0 人站在隊伍 最前方 ，第 (n - 1) 人站在隊伍 最後方 。給你一個下標從 0 開始的整數陣列 tickets ，陣列長度為 n ，其中第 i 人想要購買的票數為 tickets[i] 。
每個人買票都需要用掉 恰好 1 秒 。一個人 一次只能買一張票 ，如果需要購買更多票，他必須走到 隊尾 重新排隊（瞬間 發生，不計時間）。如果一個人沒有剩下需要買的票，那他將會 離開 隊伍。
返回位於位置 k（下標從 0 開始）的人完成買票需要的時間（以秒為單位）。
思路
這道題的暴力解是用一個佇列，把所有人需要購買的票數放進去，然後開始遍歷每一個元素。但是如果人數太多（n 太大）就一定會超時。
這裡我提供一個次優解，不需要用佇列，就是簡單的模擬。當第 k 個人還未完成購票的時候，就一直遍歷整個 input 陣列。因為每個人一次只能購買一張票，一次購買的動作耗時一秒鐘，所以我們一直遍歷整個陣列直到第 k 個人完成購票，才把迴圈 break，返回結果。
這裡同時我提供一個網上看到的最優解，時間複雜度為O(n)，但是很難想到。
複雜度
時間O(n * k)
空間O(1)
程式碼
Java實現
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;
}
}