Best Time to Buy and Sell Stock

虽然解出来了，但是不是最优的方法

Say you have an array for which the *i*th
 element is the price of a given stock on day *i*.
Design an algorithm to find the maximum profit. You may complete 
at most *two* transactions.
**Note:**You may not engage in multiple transactions at the same 
time (ie, you must sell the stock before you buy again).

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class Solution(object):
    """
    :type prices: List[int]
    :rtype: int
    """
    def maxProfit(self, prices):
        if len(prices) == 0 or len(prices) == 1:
            return 0
        F =[0] #当前效益最大值
        min=prices[0] #最小值
        Max = 0
        n=len(prices)
        for i in range(n-1):
            F.append(F[i] if F[i] > prices[i+1]-min else prices[i+1]-min)
            min = min if prices[i+1] > min else prices[i+1]
        T = [0]
        max=prices[n-1] #最大值
        for i in range(1,n):
            T.append(T[i-1] if T[i-1] > max-prices[n-i-1] else  max-prices[n-i-1] )
            max = max if prices[n-i-1] < max else prices[n-i-1]
        for i in range(0,n):
            Max = Max if Max > T[n-1-i]+F[i] else T[n-1-i]+F[i]
        return Max

看下面的解法

#前i次 交易j次的利润
f[j][i] = max(f[j][i-1], prices[i] +maxtmp);
#最大临时值
maxtmp=max（maxtmp，f[i][j - 1] - price[i]）;

class Solution {  
public:  
    int maxProfit(vector<int>& prices) {  
        if (prices.size() < 2)   
            return 0;  
        vector< vector<int> > f(3, vector<int>(prices.size(), 0));  
        for (int j = 1; j <= 2; j++) {  
            int tmpMax = f[j-1][0] - prices[0];  
            for (int i = 1; i < prices.size(); i++) {  
                f[j][i] = max(f[j][i-1], prices[i] + tmpMax);  
                tmpMax = max(tmpMax, f[j-1][i] - prices[i]);  
            }  
        }  
        return f[2][prices.size()-1];  
    }  
};