123、Best Time to Buy and Sell Stock III

Say you have an array for which the ithelement is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most twotransactions.

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.

思路

具体解法参照本题系列之 iv，将k设为2即可。

c++11

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (2 >= prices.size()/2) {
            int res = 0;
            for (int i = 1; i < prices.size(); i++)
                if (prices[i] > prices[i-1]) 
                    res += std::max(prices[i] - prices[i-1], 0);
            return res;
        }

        int hold[2+1];
        int sell[2+1];
        for (int i=0; i<=2; ++i){
            hold[i] = INT_MIN;
            sell[i] = 0;
        }

        for (int i = 0; i < prices.size(); i++) {
            for (int j = 1; j <= 2; j++) {   
                sell[j] = std::max(sell[j], hold[j] + prices[i]);
                hold[j] = std::max(hold[j], sell[j-1] - prices[i]);

            }
        }
        return sell[2];
    }
};

AC结果：

Runtime: 16 ms, faster than 25.97% of C++ online submissions for Best Time to Buy and Sell Stock III.
Memory Usage: 9.5 MB, less than 72.36% of C++ online submissions for Best Time to Buy and Sell Stock III.