LeetCode\EPI "Best Time to Buy and Sell Stock"

Also the very first problem on EPI. 

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        size_t len = prices.size();
        if (len <= 1) return 0;
        else if (len == 2)
        {
            if (prices[0] >= prices[1]) return 0;
            else return prices[1] - prices[0];
        }
        int len1 = len / 2;
        int len2 = len - len1;

        vector<int> v1; v1.assign(prices.begin(), prices.begin() + len1);
        int prof1 = maxProfit(v1);
        vector<int> v2; v2.assign(prices.begin() + len1, prices.end());
        int prof2 = maxProfit(v2);
        auto i1 = std::min_element(v1.begin(), v1.end());
        int min1 = *i1;
        auto i2 = std::max_element(v2.begin(), v2.end());
        int max1 = *i2;
        int prof3 = max1 - min1;
        return std::max(std::max(prof1, prof2), prof3);
    }
};

 Update Aug. 18:

Actually there's a O(n) algorthm. Since the logic should be: the current day's max profit is price@dayI - minPrice of day[0..i-1]. Then, a linear scan should work:

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        size_t len = prices.size();
        if (len <= 1) return 0;
        
        int maxProfit = 0;
        vector<int> dp; dp.push_back(0);
        int lowest = prices[0];
        for (int i = 1; i < prices.size(); i++)
        {
            int currP = prices[i] - lowest;
            if (currP > maxProfit) maxProfit = currP;
            if (prices[i] < lowest) lowest = prices[i];
        }
        return maxProfit;
    }
};