leetCode -- Best Time to Buy and Sell Stock II [贪心算法]

Best Time to Buy and Sell Stock II

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

   

 我的实现方法（贪心）：

 

'''

Created on Nov 13, 2014



@author: ScottGu<[email protected], [email protected]>

'''

class Solution:

    # @param prices, a list of integer

    # @return an integer

    def maxProfit(self, prices):

        self.__init__()

        for i in range(len(prices)):

            prices[i]=prices[i]+1

        prices.append(0)

        

        self.trade(prices)

        return self.profit

    

    def __init__(self):

        self.profit=0

        self.bought=0

        

    def trade(self, prices):

        if(prices==None): return 

            

        for i in range(1, len(prices)-1):

            if (prices[i-1] <prices[i] >=prices[i+1]): #sell 

                if(self.bought==0): self.bought=prices[i-1]

                self.profit+=(prices[i]-self.bought)

                self.bought=0

            

            if (prices[i-1] >=prices[i] <prices[i+1]):#buy

                self.bought=prices[i]

                

            if (prices[i-1] <prices[i] <prices[i+1]): #maybe buy

                if(self.bought==0): self.bought=prices[i-1]

        if(self.bought>0):

            self.profit+=(prices[-1]-self.bought)

            

if __name__ == '__main__':

    so=Solution() 

    

    prices=[1,2,3,4,5,3,3,3,2,6,7,3,4]

    print prices

    print so.maxProfit(prices)

        

    prices=[1,2]

    print prices

    print so.maxProfit(prices)

            

    prices=[2,2,5]

    print prices

    print so.maxProfit(prices)

        

    prices=[2,1,2,0,1]

    print prices

    print so.maxProfit(prices)

    prices=[2,1,2,1,0,1,2]

    print prices

    print so.maxProfit(prices)

    

    prices=    [0,5,5,6,2,1,1,3]

    print prices

    print so.maxProfit(prices)

    pass

 

 贪心算法的特点是一条路走到黑，把问题分解成若干子问题，逐个解决，问题集越来越小直到全解完，这时结果集就认为是最优解。

但贪心算法并不能在所有场景下确保结果是最优解，在一些情况下结果是次优解，看这个问题：

　　假如某个国家只有1元、5元和11元面值的钞票，这时如果有商人要【找零15元】，问最少钞票张数？

　　假如使用贪心算法，则结果为一张11元和4张1元钞票，共5张。而实际正确结果应该为3张5元钞票。

 

那么问题来了，在什么场景下使用贪心算法能获得最优解？

答：局部最优解能决定全局最优解。简单地说，问题能够分解成子问题来解决，子问题的最优解能递推到最终问题的最优解。

 

贪心法一般不能得到我们所要求的答案。一旦一个问题可以通过贪心法来解决，那么贪心法一般是解决这个问题的最好办法。由于贪心法的高效性以及其所求得的答案比较接近最优结果，贪心法也可以用作辅助算法或者直接解决一些要求结果不特别精确的问题。