Say you have an array for which the number i 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.
Example：
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.
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.
Note：
You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
解释下题目：
做两笔交易，获得最大的收益
1. 动态规划
实际耗时：3ms
public int maxProfit(int[] prices) {
int buy1 = Integer.MIN_VALUE, buy2 = Integer.MIN_VALUE;
int sell1 = 0, sell2 = 0;
for (int i : prices) {
//如果我们进行了第二次出售可以获得的总钱数
sell2 = Math.max(sell2, buy2 + i);
//如果我们进行了第二次购入可以获得的总钱数
buy2 = Math.max(buy2, sell1 - i);
//---------------分割线---------------------
//如果我们进行了第一次出售可以获得的总钱数
sell1 = Math.max(sell1, buy1 + i);
//如果我们进行了第一次购入可以获得的总钱数
buy1 = Math.max(buy1, -i);
}
return sell2;
}
思路在注释里，很好懂
