Alice and Bob want to water nplants in their garden. The plants are arranged in a row and are labeled from 0to n - 1from left to right where the ithplant is located at x = i.

(资料图片仅供参考)

Each plant needs a specific amount of water. Alice and Bob have a watering can each, initially full. They water the plants in the following way:

Alice waters the plants in order from left to right, starting from the 0thplant. Bob waters the plants in order from right to left, starting from the (n - 1)thplant. They begin watering the plants simultaneously.

It takes the same amount of time to water each plant regardless of how much water it needs.

Alice/Bob must water the plant if they have enough in their can to fully water it. Otherwise, they first refill their can (instantaneously) then water the plant.

In case both Alice and Bob reach the same plant, the one with more water currently in his/her watering can should water this plant. If they have the same amount of water, then Alice should water this plant.

Given a 0-indexed integer array plantsof nintegers, where plants[i]is the amount of water the ithplant needs, and two integers capacityAand capacityBrepresenting the capacities of Alice's and Bob's watering cans respectively, return the number of times they have to refill to water all the plants.

Example 1:

Input: 

plants = [2,2,3,3], capacityA = 5, capacityB = 5

Output: 1

Explanation:

- Initially, Alice and Bob have 5 units of water each in their watering cans.

- Alice waters plant 0, Bob waters plant 3.

- Alice and Bob now have 3 units and 2 units of water respectively. 

- Alice has enough water for plant 1, so she waters it. Bob does not have enough water for plant 2, so he refills his can then waters it.So, the total number of times they have to refill to water all the plants is 0 + 0 + 1 + 0 = 1.

Example 2:

Input: plants = [2,2,3,3], capacityA = 3, capacityB = 4

Output: 2

Explanation:

- Initially, Alice and Bob have 3 units and 4 units of water in their watering cans respectively.- Alice waters plant 0, Bob waters plant 3.

- Alice and Bob now have 1 unit of water each, and need to water plants 1 and 2 respectively.

- Since neither of them have enough water for their current plants, they refill their cans and then water the plants.

So, the total number of times they have to refill to water all the plants is 0 + 1 + 1 + 0 = 2.

Example 3:

Input: 

plants = [5], capacityA = 10, capacityB = 8

Output: 0

Explanation:

- There is only one plant. 

- Alice's watering can has 10 units of water, whereas Bob's can has 8 units. 

Since Alice has more water in her can, she waters this plant.So, the total number of times they have to refill is 0.

Constraints:

n == plants.length

1 <= n <= 105

1 <= plants[i] <= 106

max(plants[i]) <= capacityA, capacityB <= 109

主要是判断2种情况，1种是plant的长度是奇数，一种是偶数的情况，

分别去看，然后写两个函数，分别是从2端计算，需要蓄水的次数跟剩下的水，

最后根据题意去比对即可；

下面是代码：没想到速度会这么快。

Runtime: 4 ms, faster than 99.18% of Java online submissions for Watering Plants II.

Memory Usage: 58.9 MB, less than 55.74% of Java online submissions for Watering Plants II.