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 * 10000
0 <= apples[i], days[i] <= 2 * 10000
days[i] = 0 if and only if apples[i] = 0.

题目大意 #

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。

解题思路 #

贪心算法和最小堆

data中的end表示腐烂的日期，left表示拥有的苹果数量
贪心:每天吃掉end最小但没有腐烂的苹果
最小堆:构造类型为数组(数组中元素的类型为data)的最小堆

代码 #

package leetcode

import "container/heap"

func eatenApples(apples []int, days []int) int {
	h := hp{}
	i := 0
	var ans int
	for ; i < len(apples); i++ {
		for len(h) > 0 && h[0].end <= i {
			heap.Pop(&h)
		}
		if apples[i] > 0 {
			heap.Push(&h, data{apples[i], i + days[i]})
		}
		if len(h) > 0 {
			minData := heap.Pop(&h).(data)
			ans++
			if minData.left > 1 {
				heap.Push(&h, data{minData.left - 1, minData.end})
			}
		}
	}
	for len(h) > 0 {
		for len(h) > 0 && h[0].end <= i {
			heap.Pop(&h)
		}
		if len(h) == 0 {
			break
		}
		minData := heap.Pop(&h).(data)
		nums := min(minData.left, minData.end-i)
		ans += nums
		i += nums
	}
	return ans
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

type data struct {
	left int
	end  int
}

type hp []data

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *hp) Push(x interface{}) {
	*h = append(*h, x.(data))
}

func (h *hp) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

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