题目描述

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).

解法一

思路

注意这题对交易的次数没有限制。既然开启了“上帝视角”,那么对于每一天来说,如果明天股票涨了,那就今天买进,明天卖掉;如果后天涨了,那明天买进,后天卖掉,如此往复即可。

Python

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        profit = 0
        for i in range(1, len(prices)):
            gain = prices[i] - prices[i - 1]
            if gain > 0:
                profit += gain
        return profit

Java

public class Solution {
    public int maxProfit(int[] prices) {
        int profit = 0;
        for (int i = 1; i < prices.length; i++) {
            int gain = prices[i] - prices[i - 1];
            if (gain > 0) {
                profit += gain;
            }
        }
        return profit;
    }
}

C++

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