Best Time to Buy and Sell Stock III (H)

題目

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 at most two transactions.

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: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 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.

題意

在一個時刻買入股票，在另一個時刻賣出股票，最多可進行2次交易，求能得到的最大收益。

思路

和 0121 很像，只是限定了最多兩次交易。可以在0121的基礎上進行改進：將陣列分成左右兩個子陣列A(1 -> k)和B(k+1 -> n)，分別找到最大收益並相加，改變k找到最大的和就是答案。但這樣做有一個問題，即沒有考慮A中買入B中賣出的情況。繼續改進，我們給A中的最大收益加上一個限制，即必須在第k天賣出，這樣第二次交易就能完全落在B中。

程式碼實現

Java

class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length == 0) {
            return 0;
        }

        int profit = 0;
        int minLeft = prices[0];
        int maxRight = prices[prices.length - 1];
        int[] pRightMax = new int[prices.length];

        // 找到從i開始的右區間能得到的最大收益
        for (int i = prices.length - 2; i >= 0; i--) {
            pRightMax[i] = Math.max(pRightMax[i + 1], maxRight - prices[i]);
            maxRight = Math.max(maxRight, prices[i]);
        }

        for (int i = 1; i < prices.length; i++) {
            // prices[i]-minLeft 為左區間在第i天賣出能得到的最大收益
            if (prices[i] - minLeft >= 0) {
                profit = Math.max(profit, prices[i] - minLeft + (i == prices.length - 1 ? 0 : pRightMax[i + 1]));
            }
            minLeft = Math.min(minLeft, prices[i]);
        }

        return profit;
    }
}