Check LintCode 1000. Best Time to Buy and Sell Stock with Transaction Fee

class Solution {
public:
    /**
     * @param prices: a list of integers
     * @param fee: a integer
     * @return: return a integer
     */
    int maxProfit(vector<int> &prices, int fee) {
        // write your code here
        // 双数组DP思想的再次使用：
        // f[i]状态表示: 前i个元素最大收益。stay at i-th element:
        //1, sell: f[i] = ? + prices[i - 1] - fee. This ? state needs at i-1 th day, have the stock held. This state is the maxProfit that i-1 th day stock was hold. Denote ? as P[i].
        //2, not sell: f[i] = f[i - 1];
        // now P状态的转移，情况：
        // 1，继续hold P[i] = P[i - 1]
        // 2，买入 P[i] = f[i - 1] - prices[i - 1].
        //初始: f[0] = 0;  P[1] = -prices[0]. P[1] = max(P[0), f[0] - prices[0])。所以P[0] = INT_MIN;
        //结果: f[n].
        int sizeP = prices.size();
        if(sizeP <= 1){
            return 0;
        }
        vector<int> global(sizeP + 1, 0);
        vector<int> local(sizeP + 1, 0);
        //Initial
        global[0] = 0;
        local[0] = -prices[0];
        for(int i = 1; i <= sizeP; i++){
            global[i] = max(global[i - 1], local[i - 1] + prices[i - 1] - fee);
            local[i] = max(local[i -1], global[i - 1] - prices[i - 1]);
        }
        return global[sizeP];
    }
};