Leetcode第1011题 在 D 天内送达包裹的能力C++解法

主要在两个地方，第一是low和high的值，low应该为数组元素最大值（如果low小于数组最大值，那么将由元素无法装车），high为数组和（D最少为1的时候）。
再有辅助函数要写好，两种思路，一种是累加一种是累减。累加可以看当重量为m时，需要几天才能运完。累减可以看重量为m，D天内是否能运完

class Solution {
    
    
public:
    int shipWithinDays(vector<int>& weights, int D) {
    
    
        int low=0,high=0;
        for(int p:weights) {
    
    low=max(low,p);high+=p;}
        if(D==1)    return high;
        while(low<high)
        {
    
    
            int mid=low+(high-low)/2;
            if(calTime(weights,mid)<=D)
                high=mid;
            else
                low=mid+1;
        }
        return high;
    }
    int calTime(vector<int>& weights,int m)
    {
    
    
        int n=weights.size(),res=0;
        for(int i=0;i<n;++res)
        {
    
    
            int sum_m=0;
            while(i<n){
    
    
                sum_m+=weights[i];
                if(sum_m<=m)
                i++;
                else break;
            }
        }
        return res;
    }

};

累减做法

class Solution {
    
    
public:
    int shipWithinDays(vector<int>& weights, int D) {
    
    
        int low=0,high=0;
        for(int p:weights) {
    
    low=max(low,p);high+=p;}
        if(D==1)    return high;
        while(low<high)
        {
    
    
            int mid=low+(high-low)/2;
            if(calTime(weights,mid,D))
                high=mid;
            else
                low=mid+1;
        }
        return high;
    }
    bool calTime(vector<int>& weights,int& m,int& D)
    {
    
    
        int n=weights.size();
        for(int t=0,i=0;t<D;++t)
        {
    
    
            int sav=m;
            while(i<n&&(sav-weights[i])>=0)
                sav-=weights[i++];
            if(i==n)
            return true;            
        }
        return false;
    }

};