[leetcode] 122. Best Time to Buy and Sell Stock II

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分类专栏： C++ leetcode题解文章标签： leetcode 算法职场和发展

于 2019-01-20 20:33:10 首次发布

本文链接：https://blog.youkuaiyun.com/w5688414/article/details/86565777

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本文介绍了一种算法，用于计算给定股票价格序列时的最大可能利润。算法允许进行多次买入卖出操作，但不能同时持有多个股票。通过分析股价趋势，当股价上升时进行交易以获取利润。提供了C++和Python两种语言的实现代码。

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Description

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 (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

分析

题目的意思为：设计一个算法求总的最大的收益，可以多次交易，但是一次交易完后，才能进行下一次交易

如果股价呈现递增的趋势，就可以盈利，然后就可以交易。这样的话，我们比较一下如果价格比上一次价格高的话，就直接交易，然后把盈利求和就是总的收益。

代码

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

Python实现

这道题还可以用动态规划来实现，贪心算法不容易想出来，但动态规划可以进行推导得到。dp[i][0]表示第i天交易完以后，手里没有股票的最大利润；dp[i][1]表示第i天交易完以后，手里还有1支股票的最大利润，推导公式：

# 保持不变 or 卖了
dp[i][0] =max(dp[i-1][0],dp[i-1][1]+prices[i])
# 保持不变 or 持有当前的股票
dp[i][1] = max(dp[i-1][1],dp[i-1][0]-prices[i])

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        dp = [[0]*2 for i in range(n)]
        dp[0][0]=0
        dp[0][1]=-prices[0]

        for i in range(1,n):
            dp[i][0] =max(dp[i-1][0],dp[i-1][1]+prices[i])
            dp[i][1] = max(dp[i-1][1],dp[i-1][0]-prices[i])
        return dp[n-1][0]