leetcode121

public class Solution {
    public int MaxProfit(int[] prices)
        {
            //寻找最优极值点（包括2个端点)

            if (prices.Length < )
            {
                return ;
            }
            else if (prices.Length == )
            {
                return prices[] - prices[] >  ? prices[] - prices[] : ;
            }
            else//至少3个点
            {
                //先对原始数据进行压缩
                var list = new List<int>();
                for (int i = ; i < prices.Length - ; i++)
                {
                    if (prices[i] != prices[i + ])
                    {
                        list.Add(prices[i]);
                    }
                }
                var last = list.LastOrDefault();
                if (last != prices[prices.Length - ])
                {
                    list.Add(prices[prices.Length - ]);
                }

                var dic = new Dictionary<int, int>();//记录所有极值点极其类型
                //Key是index，Value是类型：-1是极小，1是极大

                //list已经压缩，没有平行点了
                for (int i = ; i < list.Count - ; i++)
                {
                    //判断是否是极大值点
                    if (list[i - ] < list[i] && list[i] > list[i + ])
                    {
                        dic.Add(i, );
                    }
                    else if (list[i - ] > list[i] && list[i] < list[i + ])
                    {
                        dic.Add(i, -);
                    }
                    //判断是否是极小值点
                }

                //处理端点
                if (dic.Count == )
                {
                    return list[list.Count - ] - list[] >  ? list[list.Count - ] - list[] : ;
                }
                else
                {
                    var list2 = dic.OrderBy(x => x.Key).ToList();
                    var d1 = list2.FirstOrDefault();
                    var d2 = list2.LastOrDefault();
                    if (d1.Value == )
                    {
                        list2.Add(new KeyValuePair<int, int>(, -));
                    }
                    else
                    {
                        list2.Add(new KeyValuePair<int, int>(, ));
                    }

                    if (d2.Value == )
                    {
                        list2.Add(new KeyValuePair<int, int>(list.Count - , -));
                    }
                    else
                    {
                        list2.Add(new KeyValuePair<int, int>(list.Count - , ));
                    }

                    list2 = list2.OrderBy(x => x.Key).ToList();//得到全部的极值点

                    var maxProfit = ;

                    for (int i = ; i < list2.Count; i++)
                    {
                        if (list2[i].Value == -)
                        {
                            for (int j = i; j < list2.Count; j++)
                            {
                                if (list2[j].Value == )
                                {
                                    if (list[list2[j].Key] - list[list2[i].Key] > maxProfit)
                                    {
                                        maxProfit = list[list2[j].Key] - list[list2[i].Key];
                                    }
                                }
                            }
                        }
                    }
                    return maxProfit;
                }

            }
        }
}

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/#/description

原来的计算方式太复杂了，使用优化的方式如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int minprice = INT_MAX;
        int maxprofit = ;
        for (int i = ; i < prices.size(); i++)
        {
            if (minprice > prices[i])
            {
                minprice = prices[i];
            }
            else if (prices[i] - minprice > maxprofit)
            {
                maxprofit = prices[i] - minprice;
            }
        }
        return maxprofit;
    }
};

补充一个python的实现：

 import sys
 class Solution:
     def maxProfit(self, prices: List[int]) -> int:
         n = len(prices)
         maxval =
         minval = sys.maxsize
         for i in range(n):
             cur = prices[i]
             minval = min(minval,cur)
             maxval = max(maxval,cur-minval)
         return maxval