leetcode 121. Best Time to Buy and Sell Stock 、122.Best Time to Buy and Sell Stock II 、309. Best Time to Buy and Sell Stock with Cooldown 、714. Best Time to Buy and Sell Stock with Transaction Fee

121. Best Time to Buy and Sell Stock

题目的要求是只买卖一次，买的价格越低，卖的价格越高，肯定收益就越大

遍历整个数组，维护一个当前位置之前最低的买入价格，然后每次计算当前位置价格与之前最低价格的差值，获得最大差值即为结果

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

  

122.Best Time to Buy and Sell Stock II

可以买任意次，所以只要当前的价格比之前的一个多，就买卖，这样最多

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

309. Best Time to Buy and Sell Stock with Cooldown

https://www.cnblogs.com/grandyang/p/4997417.html

https://blog.csdn.net/qq508618087/article/details/51671504

注意题目的注释：you must sell the stock before you buy again，也就是说不能买了又买，卖了又卖，只能买一次再卖一次

用buy、sell两个数组表示当前位置以buy、sell操作结尾的最大收入，以buy为例，当前状态只可能来自两种情况：一、这一位置不做任何操作，就是不买也不卖  二、这一位置买，那之前的状态就是i-2的时候卖出了东西

初始化的时候注意0、1都要初始化，因为i-2，从2开始的循环。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int length = prices.size();
        if(length <= 0)
            return 0;
        vector<int> buy(length+1,0);
        vector<int> sell(length+1,0);
        buy[1] = -prices[0];
        for(int i = 2;i <= length;i++){
            buy[i] = max(buy[i-1],sell[i-2] - prices[i-1]);
            sell[i]  = max(sell[i-1],buy[i-1] + prices[i-1]);
        }
        return max(buy[length],sell[length]);
    }
};

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

https://www.cnblogs.com/grandyang/p/7776979.html

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        vector<int> buy(prices.size(),0);
        vector<int> sell(prices.size(),0);
        buy[0] = -prices[0];
        for(int i = 1;i < prices.size();i++){
            buy[i] = max(buy[i-1],sell[i-1] - prices[i]);
            sell[i] = max(sell[i-1],buy[i-1] + prices[i] - fee);
        }
        return max(buy[prices.size() - 1],sell[prices.size() - 1]);
    }
};