2073. Time Needed to Buy Tickets

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2073. Time Needed to Buy Tickets

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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

难度 : Easy

思路1

暴力

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

思路2

可以把数组分成2部分 [0, k] 和 [k+1, n) 假设数组长度为n
对于区间[0,k] 每个人i 购买票需要的时间是 min(tickets[i], tickets[k])
对于区间[k+1 ,n) 每个人 i 购买票需要的时间是 min(tickets[i], tickets[k] - 1)
因为当他买够了票之后就停止计时了，所以之前的 tickets[k] - 1 轮，k后面的人如果买票的话，时间会加到结果中

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class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
        int ans = 0;        
        for (int i = 0; i < tickets.length; i++) {
            if (i <= k) {
                ans += Math.min(tickets[i], tickets[k]);
            } else {
                ans += Math.min(tickets[k] - 1, tickets[i]);
            }
        }
        return ans;
    }
}