LeetCode1011.Capacity To Ship Packages Within D Days（在 D 天内送达包裹的能力）

Description

A conveyor belt has packages that must be shipped from one port to another within D days.

The i-th package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.

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传送带上的包裹必须在 D 天内从一个港口运送到另一个港口。

传送带上的第 i 个包裹的重量为 weights[i]。每一天，我们都会按给出重量的顺序往传送带上装载包裹。我们装载的重量不会超过船的最大运载重量。

返回能在 D 天内将传送带上的所有包裹送达的船的最低运载能力。

题目链接：https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/

个人主页：http://redtongue.cn or https://redtongue.github.io/

Difficulty: medium

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
Output: 15
Explanation: 
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10



Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed. 

Example 2:

Input: weights = [3,2,2,4,1,4], D = 3
Output: 6
Explanation: 
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], D = 4
Output: 3
Explanation: 
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Note:

• 1 <= D <= weights.length <= 50000
• 1 <= weights[i] <= 500

分析

• 若对于最大容量每次增一，时间复杂度是 O(NK)O(NK),时间会超限，利用二分查找的方法，时间复杂度可以降到 O(KlogN)O(KlogN)，N是capacity的变换区间，K是包裹数;
• 初始capacity的最小值是最大包裹重量，最大值是包裹重量的总和，逐次遍历weights，更新left，mid，right，直到left与right重合；
• 返回left。

参考代码

def shipWithinDays(self, weights, D):
    left, right = max(weights), sum(weights)
    while left < right:
        mid, need, cur = (left + right) / 2, 1, 0
        for w in weights:
            if cur + w > mid:
                need += 1
                cur = 0
            cur += w
        if need > D: left = mid + 1
        else: right = mid
    return left