123 Best Time to Buy and Sell Stock III

123. Best Time to Buy and Sell Stock III

1. Question

Say you have an array for which thei_thelement 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.

2. Implementation

(1) DP

思路: 188的简化版，参考188即可

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) {
            return 0;
        }

        int k = 2;
        int[][] dp = new int[k + 1][prices.length];

        for (int i = 1; i <= k; i++) {
            int maxDiff = -prices[0];
            for (int j = 1; j < prices.length; j++) {
                dp[i][j] = Math.max(dp[i][j - 1], prices[j] + maxDiff);
                maxDiff = Math.max(maxDiff, dp[i - 1][j] - prices[j]);
            }
        }
        return dp[k][prices.length - 1];
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(n), 空间复杂度O(n)