714. Best Time to Buy and Sell Stock with Transaction Fee

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/
There could be two possible states after visiting prices[i-1]:
1. hold: there is 1 share of stock at hand. The total cash is named hold.
2. sold: there is 0 share of stock at hand. The total cash is named sold.
After visiting prices[i], the states are
1. hold = max(hold at i-1, sold at i-1 and buy 1 share at i)
2. sold = max(sold at i-1, hold at i-1 and sell 1 share at i)
    int maxProfit(vector<int>& prices, int fee) {
        // at i=0, the total cash at hand could be sold or hold:
        int sold = 0, hold = -prices[0];
        for(int i=1; i<prices.size(); i++) {
            int h = max(hold, sold - prices[i]);
            int s = max(sold, hold + prices[i] - fee);
            sold = s;
            hold = h;
        }
        return sold;
    }
or, check local min and max
    int maxProfit(vector<int>& prices, int fee) {
        prices.push_back(-50000);
        int hold = -prices[0], cash = 0;
        for(int i=1; i<prices.size()-1; i++) {
            if(prices[i]<=prices[i-1] && prices[i]<prices[i+1]) {
                // local minimum
                hold = max(hold, cash - prices[i]);
            }
            else if(prices[i]>=prices[i-1] && prices[i]>prices[i+1]) {
                // local maximum
                cash = max(cash, hold + prices[i] -fee);
            }
        }
        return cash;
    }