LeetCode --- 122. Best Time to Buy and Sell Stock II-优快云博客

本文介绍了一种在给定股票价格序列中寻找最大利润的算法。通过贪心策略，只要发现价格上升趋势，即刻买入并卖出，实现多次交易最大化利润。文章提供了C++代码示例，展示了如何在一天内完成算法实现。

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122. Best Time to Buy and Sell Stock II

Difficulty: Easy

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 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.

Solution

思路

贪心算法，如果今天比明天价格低，则今天买入，明天售出，循环一次即可，时间复杂度 $O (n)$ 。
Language: C++

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