leetcode_1014. Capacity To Ship Packages Within D Days

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=0,i=0,day=0;
        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=0;
                day++;
            }
            if(i>=weights.size() && temp<=maxn && temp>0)
                day++;
        }
        return day<=D;
    }

    int shipWithinDays(vector<int>& weights, int D)
    {
        int sum_weight=0, max_item_weight=INT_MIN;
        for(int i=0; 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=0;
        while(low<=high)
        {
            int mid = (low+high)/2;
            if(can(weights,D,mid))
            {
                res=mid;
                high = mid-1;
            }
            else
                low = mid+1;
        }
        return res;
    }
};