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

传送带要在D天内把所有货物传送完,但是传送带每天有传送容量的限制,问保证货物在D天内传送完的最小容量限制。货物重量以weights[]给出。

  1. 1 <= D <= weights.length <= 50000
  2. 1 <= weights[i] <= 500

一开始想从weights来求最小容量,发现无论如何都不能达到线性时间复杂度。想了好久,突然想到可以二分最小容量,然后对容量验证是否能达到要求。

此外,容量初始最小值为单个货物的最大重量,容量最大值为所有货物重量加和。

class Solution
{
public:
bool can(vector<int>& weights, int D, int maxn)
{
int temp=,i=,day=;
while(i<weights.size())
{
while(i<weights.size() && temp+weights[i]<=maxn)
{
temp+=weights[i];
i++;
}
if(i<weights.size()&&temp+weights[i]>maxn)
{
temp=;
day++;
}
if(i>=weights.size() && temp<=maxn && temp>)
day++;
}
return day<=D;
} int shipWithinDays(vector<int>& weights, int D)
{
int sum_weight=, max_item_weight=INT_MIN;
for(int i=; i<weights.size(); i++)
{
sum_weight+=weights[i];
max_item_weight = max(max_item_weight, weights[i]);
}
int low=max_item_weight, high=sum_weight, res=;
while(low<=high)
{
int mid = (low+high)/;
if(can(weights,D,mid))
{
res=mid;
high = mid-;
}
else
low = mid+;
}
return res;
}
};

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