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 ;
int min_num = prices[];
int profit = ;
for(int i = ;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 = ;
for(int i = ;i < prices.size();i++){
if(prices[i] > prices[i-])
res += prices[i] - prices[i-];
}
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 <= )
return ;
vector<int> buy(length+,);
vector<int> sell(length+,);
buy[] = -prices[];
for(int i = ;i <= length;i++){
buy[i] = max(buy[i-],sell[i-] - prices[i-]);
sell[i] = max(sell[i-],buy[i-] + prices[i-]);
}
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(),);
vector<int> sell(prices.size(),);
buy[] = -prices[];
for(int i = ;i < prices.size();i++){
buy[i] = max(buy[i-],sell[i-] - prices[i]);
sell[i] = max(sell[i-],buy[i-] + prices[i] - fee);
}
return max(buy[prices.size() - ],sell[prices.size() - ]);
}
};