leetcode Best Time to Buy and Sell Stock I&&II&&III

I

class Solution {
public:
    int maxProfit(vector<int> &prices) 
    {
        if(prices.size()==0||prices.size()==1)return 0;
        int min=prices[0];
        int max=prices[0];
        int diff=0;
        for(int i=1;i<prices.size();i++)
        {
            if(prices[i]<min)min=prices[i];        
            if(prices[i]-min>diff)diff=prices[i]-min;         
        }
        return diff;
    }
};

II

class Solution { 
public: 
    int maxProfit(vector<int> &prices)  
    { 
        if(prices.size()==0||prices.size()==1)return 0; 
        int max=prices[prices.size()-1];
        int min=prices[prices.size()-1];
        int maxprofit=0;
        for(int i=prices.size()-2;i>=0;i--) 
        {
            if(prices[i]<min)
            {
                min=prices[i];
            }
            else
            {
               maxprofit+=max-min;
               max=min=prices[i];
            }
        } 
        maxprofit+=max-min;
        return maxprofit; 
    } 
};

III

class Solution {
public:
int maxProfit(vector<int> &prices) {
        int size = prices.size();
        if (size == 0)
            return 0;
        vector<int> f1(size);
        vector<int> f2(size);
        int minV = prices[0];
        for (int i = 1; i < size; i++){
            minV = std::min(minV, prices[i]);
            f1[i] = std::max(f1[i-1], prices[i] - minV);
        }
        int maxV = prices[size-1];
        f2[size-1] = 0;
        for (int i = size-2; i >= 0; i--){
            maxV = std::max(maxV, prices[i]);
            f2[i] = std::max(f2[i+1], maxV - prices[i]);
        }
        int sum = 0;
        for (int i = 0; i < size; i++)
            sum = std::max(sum, f1[i] + f2[i]);
        return sum;
    }
};

对于II，即是求所有递增子序列的和，对于III，用两个数组，一个从前往后加，一个从后往前加，求最大值即可。