[LeetCode] 123. Best Time to Buy and Sell Stock III

[frdmu]

You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

 

Example 1:

Input: prices = [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: prices = [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: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Example 4:

Input: prices = [1]
Output: 0

Constraints:

• 1 <= prices.length <= 10^5
• 0 <= prices[i] <= 10^5

 

解法一：
动态规划。一天有五个状态，代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(5, 0));
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];

        for (int i = 1; i < n; i++) {
            dp[i][0] = dp[i-1][0];
            dp[i][1] = max(dp[i-1][1], dp[i-1][0]-prices[i]);
            dp[i][2] = max(dp[i-1][2], dp[i-1][1]+prices[i]);
            dp[i][3] = max(dp[i-1][3], dp[i-1][2]-prices[i]);
            dp[i][4] = max(dp[i-1][4], dp[i-1][3]+prices[i]); 
        }

        return dp[n-1][4];
    }
};

Refer:
123.买卖股票的最佳时机III

解法二：
动态规划。使用统一框架推导动态规划总结:股票买卖问题，代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        // write code here
        int n = prices.size();
        if(n==0) return 0;
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(3, vector<int>(2, 0)));
     
        for (int i = 0; i < n; i++) {
            dp[i][0][0] = 0;
            dp[i][0][1] = INT_MIN;
        }
        for (int k = 0; k < 3; k++) {
            dp[0][k][0] = 0;
            dp[0][k][1] = -prices[0];
        }
        for (int i = 1; i < n; i++) {
            for (int k = 1; k <= 2; k++) {
                dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1]+prices[i]);
                dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0]-prices[i]);
            }
        }
        
        return dp[n-1][2][0];
    }
};
/*
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1]+prices[i])
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0]-prices[i])

base case:
dp[-1][k][0] = dp[i][0][0] = 0;
dp[-1][k][1] = dp[i][0][1] = -inf;
*/

Refer:
团灭 LeetCode 股票买卖问题