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

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

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: 7Explanation: 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: 4Explanation: 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: 0Explanation: 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& prices) { int sum=0; for(int i=1;iprices[i-1]){ sum+=prices[i]-prices[i-1]; } } return sum; }};

参考文献

​​[编程题]best-time-to-buy-and-sell-stock-ii​​

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