[Leetcode 122]买股票II Best Time to Buy and Sell Stock II-优快云博客

本文介绍了一种算法，用于计算给定股票价格数组时的最大可能利润。算法允许进行多次买入和卖出操作，但不能同时持有多个股票。通过遍历数组并累计每次价格上涨的差额，实现了最大利润的计算。

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【题目】

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).

【思路】

注意数组长度

类似：https://www.cnblogs.com/inku/p/9908287.html

交易多次/交易一次

【代码】

public int maxProfit(int[] prices) {
　　int len=prices.length;
　　int win=0;
　　for(int i=1;i<len;i++){
　　　　if(prices[i-1]<prices[i]){
　　　　win+=prices[i]-prices[i-1];
　　　　}
　　}
　　return win;
}

【举例】

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.

转载于:https://www.cnblogs.com/inku/p/9933066.html