問題

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).

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.

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

自分の解答

public int maxProfit(int[] prices) {
    int maxProfit = 0;
    int buyPrice = -1;
    for (int i = 0; i < prices.length; i++) {
        int price = prices[i];
        if (buyPrice < 0) {
            buyPrice = price;
            continue;
        }
        if (buyPrice > price) {
            buyPrice = price;
            continue;
        }
        if (buyPrice < price && (i + 1 == prices.length || price >= prices[i + 1])) {
            maxProfit += (price - buyPrice);
            buyPrice = -1;
        }
    }
    return maxProfit;
}

コード理解

別解１

class Solution {
    public int maxProfit(int[] prices) {
        int i = 0;
        int valley = prices[0];
        int peak = prices[0];
        int maxprofit = 0;
        while (i < prices.length - 1) {
            while (i < prices.length - 1 && prices[i] >= prices[i + 1])
                i++;
            valley = prices[i];
            while (i < prices.length - 1 && prices[i] <= prices[i + 1])
                i++;
            peak = prices[i];
            maxprofit += peak - valley;
        }
        return maxprofit;
    }
}

別解２

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