Leetcode 1705. 吃苹果的最大数目(medium)

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题目

有一棵特殊的苹果树，一连 n 天，每天都可以长出若干个苹果。在第 i 天，树上会长出 apples[i] 个苹果，这些苹果将会在 days[i] 天后（也就是说，第 i + days[i] 天时）腐烂，变得无法食用。也可能有那么几天，树上不会长出新的苹果，此时用 apples[i] == 0 且 days[i] == 0 表示。

你打算每天最多吃一个苹果来保证营养均衡。注意，你可以在这 n 天之后继续吃苹果。

给你两个长度为 n 的整数数组 days 和 apples ，返回你可以吃掉的苹果的最大数目。

示例 1：

输入：apples = [1,2,3,5,2], days = [3,2,1,4,2]
输出：7
解释：你可以吃掉 7 个苹果：
- 第一天，你吃掉第一天长出来的苹果。
- 第二天，你吃掉一个第二天长出来的苹果。
- 第三天，你吃掉一个第二天长出来的苹果。过了这一天，第三天长出来的苹果就已经腐烂了。
- 第四天到第七天，你吃的都是第四天长出来的苹果。
示例 2：

输入：apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
输出：5
解释：你可以吃掉 5 个苹果：
- 第一天到第三天，你吃的都是第一天长出来的苹果。
- 第四天和第五天不吃苹果。
- 第六天和第七天，你吃的都是第六天长出来的苹果。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-number-of-eaten-apples
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解题思路

1. 原始解法

使用一个列表food来存储信息，列表的每个元素是[i + days[i], apples[i]]

前n天：

(1). 收获苹果：当某一天的apples[i]不为0时，才向food添加元素

food = []
for i in range(n):  # at day i, we first harvest applesif apples[i] != 0:food.append([i + days[i], apples[i]])

(2).对food进行排序，依据为i + days[i]：排序是为了吃最快腐烂的苹果

# then we need to sort the list based on i+days[i]
food.sort(key=lambda x: x[0])

(3).如果有苹果，则吃一个

# now we have to find out whether we have an apple to eat
if len(food) != 0:food[0][1] -= 1count += 1

(4).丢弃所有将要腐烂的苹果（以及被吃完的food[0]）

# if we ate all apples from food[0], drop it
if len(food) != 0:if food[0][1] == 0:food.pop(0)# at the end of day, we need to throw away all rotten apples
if len(food) != 0:while (food[0][0] == i + 1):food.pop(0)if len(food) == 0:break

后续：

每一天吃一个苹果，并丢弃过期的。这里我使用了for循环，所以需要先确定最后一批苹果腐烂的那一天。如果前n天吃掉了所有水果，则无需处理后续部分。

# max_day represents the last day we have an apple
if len(food) > 0:max_day = food[-1][0]
else:max_day = n

吃苹果、丢苹果与前n天类似

结果：通过66/69测试用例，之后超时。原因应该是排序的部分。

完整代码：

class Solution:def eatenApples(self, apples: List[int], days: List[int]) -> int:count = 0n = len(apples)# use a list to store all the information# each element of the list has two components  [i+days[i], apples[i]]        food = []for i in range(n): # at day i, we first harvest applesif apples[i] != 0:food.append([i + days[i], apples[i]])# then we need to sort the list based on i+days[i]food.sort(key=lambda x: x[0])# now we have to find out whether we have an apple to eatif len(food) != 0:food[0][1] -= 1count += 1# if we ate all apples from food[0], drop itif len(food) != 0:if food[0][1] == 0:food.pop(0)# at the end of day, we need to throw away all rotten applesif len(food) != 0:while (food[0][0] == i + 1):food.pop(0)if len(food) == 0:break# max_day represents the last day we have an appleif len(food) > 0:max_day = food[-1][0]else:max_day = nfor i in range(n, max_day):# if there's no appleif len(food) == 0:break# else eat an applefood[0][1] -= 1count += 1# if we ate all apples from food[0], drop itif food[0][1] == 0:food.pop(0)# at the end of day, we need to throw away all rotten applesif len(food) != 0:while (food[0][0] == i + 1):food.pop(0)if len(food) == 0:breakreturn count

2.第一次优化

使用heapq创建一个优先队列取代列表。使用queue.PriorityQueue也可以实现目的，后者只是加入了线程方面的保护，操作仍是基于heapq。

前n天：

(1). 收获苹果

import heapq     
food = []for i in range(n): # at day i, we first harvest applesif apples[i] != 0:heapq.heappush(food, [i + days[i], apples[i]])

(2).由于使用了heapq，此时我们无需再排序，直接判断能否吃一个苹果。此处另一个区别在于，我们将food[0]直接从队列中取出了，如果我们没有吃完food[0]，需要将其再加入队列。

# now we have to find out whether we have an apple to eat
if len(food) != 0:count += 1temp = heapq.heappop(food)if temp[1] > 1:# if we haven't eaten all the apples from food[0]heapq.heappush(food, [temp[0], temp[1] - 1])

(3).丢弃将要腐烂的苹果

# at the end of day, we need to throw away all rotten apples
while(len(food) > 0):if food[0][0] == i + 1:heapq.heappop(food)else:break

后续：

在后续部分的处理中，由于使用了堆，导致无法通过food[-1]的方式得到最后一批苹果腐烂的时间。所以要用while循环判定是否还有苹果剩下，并用另一个变量记录日期。（也许这本身就是更好的处理方法）

# start from day n
i = n
# if there are still some apples left
while(len(food) > 0):count += 1temp = heapq.heappop(food)if temp[1] > 1:# if we haven't eaten all the apples from food[0]heapq.heappush(food, [temp[0], temp[1] - 1])# at the end of day, we need to throw away all rotten appleswhile(len(food) > 0):if food[0][0] == i + 1:heapq.heappop(food)else:breaki += 1

结果：通过全部测试用例。用时456ms，内存使用量19.3MB。

完整代码：

class Solution:def eatenApples(self, apples: List[int], days: List[int]) -> int:count = 0n = len(apples)# use a list to store all the information# each element of the list has two components  [i+days[i], apples[i]]   # use heapqimport heapq     food = []for i in range(n):# at day i, we first harvest applesif apples[i] != 0:heapq.heappush(food, [i + days[i], apples[i]])# now we have to find out whether we have an apple to eatif len(food) != 0:count += 1temp = heapq.heappop(food)if temp[1] > 1:# if we haven't eaten all the apples from food[0]heapq.heappush(food, [temp[0], temp[1] - 1])# at the end of day, we need to throw away all rotten appleswhile(len(food) > 0):if food[0][0] == i + 1:heapq.heappop(food)else:breaki = n# if there are still some apples leftwhile(len(food) > 0):count += 1temp = heapq.heappop(food)if temp[1] > 1:# if we haven't eaten all the apples from food[0]heapq.heappush(food, [temp[0], temp[1] - 1])# at the end of day, we need to throw away all rotten appleswhile(len(food) > 0):if food[0][0] == i + 1:heapq.heappop(food)else:breaki += 1return count

3.第二次优化

通过学习题解，更改了代码的结构。主要的改动在于，不必区分前n天及之后的部分，可以使用一个while循环一并处理。

count = 0
n = len(apples)
current_day = 0# use a list to store all the information
# each element of the list has two components  [i+days[i], apples[i]]   # use heapq
import heapq     
food = []
while food or current_day < n:'''codes'''        current_day += 1

(1). 收获苹果：只在前n天访问apples和days数组

if current_day < n:# harverst applesif apples[current_day] != 0:heapq.heappush(food, [current_day + days[current_day], apples[current_day]])

(2). 吃苹果

if food:count += 1temp = heapq.heappop(food)if temp[1] > 1:# if we haven't eaten all the apples from food[0], then push it backheapq.heappush(food, [temp[0], temp[1] - 1])

(3). 丢弃将要腐烂的水果

# at the end of day, we need to throw away all rotten apples
while(len(food) > 0):if food[0][0] == current_day + 1:heapq.heappop(food)else:break

结果：通过全部测试用例，用时524ms，内存使用量19.3MB

完整代码：

class Solution:def eatenApples(self, apples: List[int], days: List[int]) -> int:count = 0n = len(apples)current_day = 0# use a list to store all the information# each element of the list has two components  [i+days[i], apples[i]]   # use heapqimport heapq     food = []while food or current_day < n:# if we have apples or we can possibly harvest applesif current_day < n:# harverst applesif apples[current_day] != 0:heapq.heappush(food, [current_day + days[current_day], apples[current_day]])# then we need to find out whether we have an apple to eatif food:count += 1temp = heapq.heappop(food)if temp[1] > 1:# if we haven't eaten all the apples from food[0], then push it backheapq.heappush(food, [temp[0], temp[1] - 1])                # at the end of day, we need to throw away all rotten appleswhile(len(food) > 0):if food[0][0] == current_day + 1:heapq.heappop(food)else:breakcurrent_day += 1return count

4. 第三次优化

将food的每个元素从列表换成了元组，同时更改了一个判断语句。

更改：

1:添加元素

heapq.heappush(food, (current_day + days[current_day], apples[current_day]))

2. 将没吃完的food[0]推回去

d, nb = heapq.heappop(food)
if nb > 1:# if we haven't eaten all the apples from food[0], then push it backheapq.heappush(food, (d, nb - 1))

结果：通过测试用例。用时360ms，内存使用量19MB。

完整代码：

class Solution:def eatenApples(self, apples: List[int], days: List[int]) -> int:count = 0n = len(apples)current_day = 0# use a list to store all the information# each element of the list has two components  [i+days[i], apples[i]]   # use heapqimport heapq     food = []while food or current_day < n:# if we have apples or we can possibly harvest applesif current_day < n and apples[current_day] != 0:# harverst apples            heapq.heappush(food, (current_day + days[current_day], apples[current_day]))# then we need to find out whether we have an apple to eatif food:count += 1d, nb = heapq.heappop(food)if nb > 1:# if we haven't eaten all the apples from food[0], then push it backheapq.heappush(food, (d, nb - 1))# at the end of day, we need to throw away all rotten appleswhile food:if food[0][0] == current_day + 1:heapq.heappop(food)else:breakcurrent_day += 1return count