D天内送达包裹能力算法-优快云博客

本文探讨了在限定天数内完成包裹递送的算法问题，提供了两种高效的解决方案：一是利用二分查找进行优化，二是采用贪心策略实现。通过Python和C++代码实例，详细解析了算法的具体实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1011. 在 D 天内送达包裹的能力

思路：二分查找 python代码如下但是比思路3快上一倍

 1 class Solution(object):
 2     def shipWithinDays(self, weights, D):
 3         r, l = 100000, 1   # 题目给定D的最大值是50000，所以r的·初始值必须大于100000
 4         while (l + 1 )!= r:
 5             mid = (r + l) // 2
 6             if self.f(mid, D, weights):
 7                 r = mid
 8             else:
 9                 l = mid
10         return r
11 
12     def f(self, x, D, weights):
13         j = 0
14         for i in range(D):
15             t = x
16             while j < len(weights) and t >= weights[j]:
17                 t -= weights[j]
18                 j += 1
19             if j == len(weights):
20                 return True
21         return False

c++的二分查找但是比思路快上9倍

 1 class Solution {
 2 public:
 3     bool C(int x,int D,vector<int>& weights){
 4         int j=0;
 5         for(int i=0;i<D;++i){
 6             int t=x;
 7             while(j<weights.size()&&t>=weights[j]){
 8                 t-=weights[j++];
 9             }
10             if(j==weights.size())return true;
11         }
12         return false;
13     }
14     int shipWithinDays(vector<int>& weights, int D) {
15         int r=100000,l=1,mid;
16         while(l+1!=r){
17             mid=(r+l)/2;
18             if(C(mid,D,weights)){
19                 r=mid;
20             }else{
21                 l=mid;
22             }
23             //cout<<l<<" "<<r<<" "<<mid<<endl;
24         }
25         return r;
26     }
27 };

思路·3：python的贪心实现

 1 class Solution(object):
 2     def shipWithinDays(self, weights, D):
 3         """
 4         :type weights: List[int]
 5         :type D: int
 6         :rtype: int
 7         """
 8         res=max(sum(weights)//D,max(weights))
 9         while True:
10             n=1
11             temp=0
12             for each in weights:
13                 if temp+each<=res:
14                     temp+=each
15                 else:
16                     temp=each
17                     n+=1
18                     if n>D:
19                         break
20             if n<=D:
21                 return res
22             else:
23                 res+=1
24         return res