題目描述(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).

題目連結

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

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.

演算法分析

設狀態f(i)表示區間[0,i]的最大利潤，g(i)表示區間[i,n-1]的最大利潤，最終答案為max{f(i)+g(i)}。

提交程式碼：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() < 2) return 0;
        const int n = prices.size();
        int max_profit = 0;
        
        vector<int> f(n, 0); // max profit between 0 , i
        vector<int> g(n, 0); // max profit between i , n
        
        for (int i = 1, cur_min = prices[0]; i < n; ++i) {
            cur_min = min(cur_min, prices[i]);
            f[i] = max(f[i - 1], prices[i] - cur_min);
        }
        
        for (int i = n - 2, cur_max = prices[n - 1]; i >= 0; --i) {
            cur_max = max(cur_max, prices[i]);
            g[i] = cur_max - prices[i];
        }
        
        for (int i = 0; i < n; ++i) {
            max_profit = max(f[i] + g[i], max_profit);
        }
        
        return max_profit;
    }
};