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描述
You have a water dispenser that can dispense cold, warm, and hot water. Every second, you can either fill up 2 cups with different types of water, or 1 cup of any type of water.
You are given a 0-indexed integer array amount of length 3 where amount[0], amount[1], and amount[2] denote the number of cold, warm, and hot water cups you need to fill respectively. Return the minimum number of seconds needed to fill up all the cups.
Example 1:
Input: amount = [1,4,2]
Output: 4
Explanation: One way to fill up the cups is:
Second 1: Fill up a cold cup and a warm cup.
Second 2: Fill up a warm cup and a hot cup.
Second 3: Fill up a warm cup and a hot cup.
Second 4: Fill up a warm cup.
It can be proven that 4 is the minimum number of seconds needed.
Example 2:
Input: amount = [5,4,4]
Output: 7
Explanation: One way to fill up the cups is:
Second 1: Fill up a cold cup, and a hot cup.
Second 2: Fill up a cold cup, and a warm cup.
Second 3: Fill up a cold cup, and a warm cup.
Second 4: Fill up a warm cup, and a hot cup.
Second 5: Fill up a cold cup, and a hot cup.
Second 6: Fill up a cold cup, and a warm cup.
Second 7: Fill up a hot cup.
Example 3:
Input: amount = [5,0,0]
Output: 5
Explanation: Every second, we fill up a cold cup.
Note:
amount.length == 3
0 <= amount[i] <= 100
解析
根据题意，有一个饮水机，可以分配冷水、温水和热水。 每一秒可以装满 2 杯不同类型的水，或 1 杯任何类型的水。给定一个长度为 3 的 0 索引整数数组 amount ，其中 amount[0], amount[1], and amount[2] 分别表示需要填充的冷水杯、温水杯和热水杯的数量。 返回填满所有杯子所需的最少秒数。
这道题考查的就是贪心思想，我们一开始肯定是尽量多次数的进行每秒装 2 杯不同类型的水，然后再进行每秒装 1 杯任何类型的水的操作。所以我们一开始对 amount 进行排序，然后循环当 amount[-2] > 0 的时候，amount[-1] 肯定大于 0 ，所以可以将他们各自的数量减一，然后重新对 amount 进行排序，直到不满足条件之后，肯定最后只剩下 amount[-1] 还有残留的杯子，我们直接加到结果里就可以了。
时间复杂度为 O(N) ，空间复杂度为 O(1) 。
解答
class Solution(object):
def fillCups(self, amount):
"""
:type amount: List[int]
:rtype: int
"""
result = 0
amount.sort()
while amount[-2] > 0:
amount[-1] -= 1
amount[-2] -= 1
result += 1
amount.sort()
result += amount[-1]
return result
运行结果
280 / 280 test cases passed.
Status: Accepted
Runtime: 42 ms
Memory Usage: 13.2 MB
原题链接
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