[LeetCode] 2105. Watering Plants II

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

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 0th plant. Bob waters the plants in order from right to left, starting from the (n - 1)th plant. 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 plants of n integers, where plants[i] is the amount of water the ith plant needs, and two integers capacityA and capacityB representing 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

给植物浇水 II。

Alice 和 Bob 打算给花园里的 n 株植物浇水。植物排成一行，从左到右进行标记，编号从 0 到 n - 1 。其中，第 i 株植物的位置是 x = i 。

每一株植物都需要浇特定量的水。Alice 和 Bob 每人有一个水罐，最初是满的 。他们按下面描述的方式完成浇水：

Alice 按 从左到右 的顺序给植物浇水，从植物 0 开始。Bob 按 从右到左 的顺序给植物浇水，从植物 n - 1 开始。他们 同时 给植物浇水。
无论需要多少水，为每株植物浇水所需的时间都是相同的。
如果 Alice/Bob 水罐中的水足以 完全 灌溉植物，他们 必须 给植物浇水。否则，他们 首先（立即）重新装满罐子，然后给植物浇水。
如果 Alice 和 Bob 到达同一株植物，那么当前水罐中水 更多 的人会给这株植物浇水。如果他俩水量相同，那么 Alice 会给这株植物浇水。
给你一个下标从 0 开始的整数数组 plants ，数组由 n 个整数组成。其中，plants[i] 为第 i 株植物需要的水量。另有两个整数 capacityA 和 capacityB 分别表示 Alice 和 Bob 水罐的容量。返回两人浇灌所有植物过程中重新灌满水罐的 次数 。

思路

我是先做了版本一2079题才来做这道题的。原本以为从一个人浇水变成两个人浇水，还是让你算需要的步数，结果这道题其实让你算的是两个人补水的次数。

既然 Alice 和 Bob 分别从花园的两端往中间走，那么这道题的总体思路还是双指针往中间逼近。无论是 Alice 还是 Bob，只要他们当前水桶里的水不能满足其当前所在位置的浇花需求，这个人就需要补水（补水次数 + 1）。

这道题最后的 corner case 是当 Alice 和 Bob 相遇的时候，如果这个位置还未浇水，需要再补水（补水次数 + 1）。不过如果他俩水量相同，那么 Alice 优先给这株植物浇水这个条件似乎没有用上。

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
    public int minimumRefill(int[] plants, int capacityA, int capacityB) {
        int n = plants.length;
        int alice = 0;
        int bob = n - 1;
        int bucketA = capacityA;
        int bucketB = capacityB;
        int res = 0;
        while (alice < bob) {
            if (bucketA < plants[alice]) {
                res++;
                bucketA = capacityA;
            }
            bucketA -= plants[alice++];

            if (bucketB < plants[bob]) {
                res++;
                bucketB = capacityB;
            }
            bucketB -= plants[bob--];
        }
        if (alice == bob && Math.max(bucketA, bucketB) < plants[alice]) {
            res++;
        }
        return res;
    }
}