1705. Maximum Number of Eaten Apples

题目：

There is a special kind of apple tree that grows apples every day for n days. On the ith day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.

You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days.

Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.

Example 1:

Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
Output: 7
Explanation: You can eat 7 apples:
- On the first day, you eat an apple that grew on the first day.
- On the second day, you eat an apple that grew on the second day.
- On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
- On the fourth to the seventh days, you eat apples that grew on the fourth day.

Example 2:

Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
Output: 5
Explanation: You can eat 5 apples:
- On the first to the third day you eat apples that grew on the first day.
- Do nothing on the fouth and fifth days.
- On the sixth and seventh days you eat apples that grew on the sixth day.

Constraints:

apples.length == n
days.length == n
1 <= n <= 2 * 10^4
0 <= apples[i], days[i] <= 2 * 10^4
days[i] = 0 if and only if apples[i] = 0.

思路：

首先这题数据量不大，只有10的4次，所以完全能够接受遍历n。我用了哈希表存苹果要过期的时间和数量，记录要过期的最大时间。然后用三个int，分别是count，记录答案，最大能吃多少；cur，如果每天不吃的话现在手上的苹果量；act，如果当前能吃并且吃了一个的情况下，手上真实的苹果量。然后从0开始循环到找到的最大天数，每天就是往cur里加苹果和减苹果，如果当前手里大于等于1，则就吃一个，在count上记录。这样看上去是不是好像不用act？但是如果apples=[1]，days=[2]，那么没有act的情况下答案是2，这其实是最后一个test case，因为我们第二天吃的那个苹果已经在第一天被吃掉了，不可能再被吃一下，这就意味着我们需要一个act来记录真实的苹果情况。但是显然，我们不能直接用cur来记录，否则前面天天被吃，后面删减苹果的时候删减的是原本成熟时的数量，会导致cur变成负数。因此我们采用act，并且在每次更新cur的时候，把cur复制给act来更新act，这样既能记录理论上苹果数量，又可以知道真实的苹果数量，这一天能不能吃到。

代码：

class Solution {
public:
int eatenApples(vector<int>& apples, vector<int>& days) {
int time=0;
unordered_map<int,int> out;
for(int i=0;i<apples.size();i++)
{
int temp=i+days[i]+1;
out[temp]+=apples[i];
time = max(time, temp);
}
int count = 0, cur = 0,act=0;
for (int i = 0; i <= time; i++)
{
if (out.count(i))
{
cur -= out[i];
act = cur;
}
if (i - 1 < apples.size())
{
cur += apples[i - 1];
act = cur;
}
if (act >= 1)
{
count += 1;
act--;
}
}
return count;
}
};

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