Leetcode(122) Best Time to Buy and Sell Stock II

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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:

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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:

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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:

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

解法

这一题可以进行多次买卖，那只要第二天的值比今天的高，就可以在今天买入赚差价，若后天的更高，可以后天再抛出，以此类推，但每次赚的都是两日价格的差价。

具体代码如下：

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