Question

Say you have an array for which the i_th 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.

Difficulty:Hard

Category:Dynamic-Programming

Analyze

思路参考博客, 在这道题目中我们使用了两个变量：

• local, 其中local[i][j]表示了在第i天达到j次交易数的时候，能够达到的最大利润。（并且最后一次交易必须发生在第i天）
• global, 其中global[i][j]表示在第i天达到j次交易的最大利润（这里不限制交易的位置）

local[i][j] = max(global[i - 1][j - 1] + max(diff, 0), local[i - 1][j] + diff)
global[i][j] = max(local[i][j], global[i - 1][j])

所以局部第j次交易的最大利润就是： 上一天的第j-1次交易的全局最大值+昨天到今天大约0的差值 Vs 上一天交易j次的局部最大利润+昨天到今天的价格差值。

• ‘global[i - 1][j - 1] + max(diff, 0)’这种情况就是，昨天交易得到的全局最大值+今天和昨天的差价，如果差价为负数，那么就加上的是0

所谓的全局最大利润：global[i][j] = max(local[i][j], global[i - 1][j])就是今天的局部最大利润，或者昨天的第‘j’次交易的最大利润（这说明昨天到今天没有利润的）

Solution

class Solution {
 public:
  int maxProfit(vector<int> &prices) {
    if (prices.empty()) return 0;
    int n = prices.size(), g[n][3] = {0}, l[n][3] = {0};
    for (int i = 1; i < prices.size(); ++i) {
      int diff = prices[i] - prices[i - 1];
      for (int j = 1; j <= 2; ++j) {
        l[i][j] = max(g[i - 1][j - 1] + max(diff, 0), l[i - 1][j] + diff);
        g[i][j] = max(l[i][j], g[i - 1][j]);
      }
    }
    return g[n - 1][2];
  }
};