123. Best Time to Buy and Sell Stock III--Hard

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.

1.思考

• 下面有两种方法进行解决：
• 方法1：
  (1) buy1 is you buy it day x cost price;
  (2) sell1 is the profit you get sell-cost price of 1;
  (3) buy2 is you buy it on day z with the profit you got from selling the first stock you bought;
  (4) sell2 is the profit of selling the second stock, you already used the profit of first stock to buy second. Hence we only need to return sell2.
• 方法2：0 - 1 buy, 1 - 1 buy 1 sell, 2 - 2 buy 1 sell, 3 - 2 buy 2 sell.
  
  

2.实现

class Solution {
public:
	int maxProfit_Solution1(vector<int>& prices) {
		int len = prices.size();
		if (len <= 1)
			return 0;

		int buy1 = INT_MAX;
		int sell1 = 0;
		int buy2 = INT_MAX;
		int sell2 = 0;

		for (int i = 0; i<len; i++)
		{
			int p = prices[i];
			buy1 = min(buy1, p);
			sell1 = max(sell1, p - buy1);
			buy2 = min(buy2, p - sell1);
			sell2 = max(sell2, p - buy2);
		}

		return sell2;		
	}

	int maxProfit_Solution2(vector<int>& prices) {
		int n = prices.size();
		int dp[2][4]; // 0 - 1 buy, 1 - 1 buy 1 sell, 2 - 2 buy 1 sell, 3 - 2 buy 2 sell
		for (int i = 0; i<4; ++i)
			dp[0][i] = INT_MIN / 2;

		for (int i = 0; i<n; ++i){
			dp[1][0] = max(dp[0][0], -prices[i]);
			dp[1][1] = max(dp[0][1], dp[0][0] + prices[i]);
			dp[1][2] = max(dp[0][2], dp[0][1] - prices[i]);
			dp[1][3] = max(dp[0][3], dp[0][2] + prices[i]);
			swap(dp[0], dp[1]);
		}

		return max(0, max(dp[0][1], dp[0][3]));
	}
};