Best time to buy and keep stock I, II, III [leetcode]

Best time to buy and buy stock I

Only one operation: Minimum value that appears before the premin record is maintained

The Code is as follows:

    int maxProfit(vector<int> &prices) {        if (prices.size() == 0) return 0;        int profit = 0;        int preMin = prices[0];        for (int i = 1; i < prices.size(); i++)        {            if (prices[i] < preMin) preMin = prices[i];            profit = max(profit, prices[i] - preMin);        }        return profit;    }

Best time to buy and stock Stock II

Unlimited operation: When ==> current price> purchase price, sell

The Code is as follows:

    int maxProfit(vector<int> &prices) {        if (prices.size() == 0) return 0;        int profit = 0;        int buy = prices[0];        for (int i = 1; i < prices.size(); i++)        {            if (buy < prices[i])                profit += prices[i] - buy;            buy = prices[i];        }        return profit;    }

Best time to buy and buy stock III

Only two operations can be performed:

The two operations cannot overlap and can be divided into two parts: 0 .. I maximum profit FST and I .. n-1 maximum profit SND

The Code is as follows:

    int maxProfit(vector<int> &prices) {        if (prices.size() == 0) return 0;        int size = prices.size();        vector<int> fst(size);        vector<int> snd(size);                int preMin = prices[0];        for (int i = 1; i < size; i++)        {            if (preMin > prices[i]) preMin = prices[i];            fst[i] = max(fst[i - 1], prices[i] - preMin);        }        int profit = fst[size - 1];        int postMax = prices[size - 1];        for (int i = size - 2; i >= 0; i--)        {            if (postMax < prices[i]) postMax = prices[i];            snd[i] = max(snd[i + 1], postMax - prices[i]);            //update profit            profit = max(profit, snd[i] + fst[i]);        }        return profit;    }

Best time to buy and keep stock I, II, III [leetcode]