1011. Capacity To Ship Packages Within D Days

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

The ith 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.

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

Constraints:

  • 1 <= D <= weights.length <= 5 * 104
  • 1 <= weights[i] <= 500

这道题说是有一条传送带在运送包裹货物,每个包裹有各自的重量,每天要把若干包裹运送到货轮上,货轮有特定的承载量,要求在给定的D天内将所有货物装上货轮,问船的最小载重量是多少。首先来分析,由于船的载重量是固定的,而包裹在传送带上又只能按照顺序上传,并不能挑拣,所以一旦加上当前包裹超过了船的载重量,则必须要放弃这个包裹,比较极端的例子就是,假如船的载重量是 50,现在船上已经装了一个重量为1的包裹,而下一个包裹重量是 50,那么这个包裹只能装在下一条船上。知道了这一点后,再来分析一下,船的载重量的范围,先来分析一下最小值,由于所有的包裹都要上船,所以最小的船载重量至少应该是最重的那个包裹,不然上不了船了,而最大的载重量就是包裹的总重量,一条船就能拉走了。所以正确的答案就在这两个边界范围之内,挨个遍历的话实在有些太不高效了,这里就要祭出二分搜索法了,当算出了中间值 mid 后,利用这个载重量去算需要多少天能运完,然后去和D做比较,如果大于D,说明需要增加载重量,否则减少载重量,最终会终止到正确的结果。具体来看代码,left 初始化为最大的包裹重量,right 初始化为所有的包裹重量总和。然后进行 while 循环,求出 mid,同时使用两个变量 cnt 和 cur,分别用来计算需要的天数,和当前货物的重量,其中 cnt 初始化为1,至少需要一天来运货物。然后遍历所有的包裹重量,每次加到 cur,若此时 cur 大于 mid 了,说明当前包裹不能加了,将 cur 重置为当前包裹重量,为下条船做准备,然后 cnt 自增1。遍历完了之后,判断若 cnt 大于D,则 left 赋值为 mid+1,否则 right 赋值为 mid,这是博主经常用的一种二分搜索的写法,可以参见博主之前的总结帖 LeetCode Binary Search Summary 二分搜索法小结,最终返回 left 即可,参见代码如下:

class Solution {
public:
    int shipWithinDays(vector<int>& weights, int D) {
        int left = *max_element(weights.begin(), weights.end()), right = accumulate(weights.begin(), weights.end(), 0);
        while (left < right) {
            int mid = left + (right - left) / 2, cnt = 1, cur = 0;
            for (int w : weights) {
                cur += w;
                if (cur > mid) {
                    cur = w;
                    ++cnt;
                }
            }
            if (cnt > D) left = mid + 1;
            else right = mid;
        }
        return left;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1011

参考资料:

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

https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/discuss/256737/C%2B%2B-Binary-Search

https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/discuss/256729/JavaC%2B%2BPython-Binary-Search

LeetCode All in One 题目讲解汇总(持续更新中…)


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