[LeetCode] 188. Best Time to Buy and Sell Stock IV 买卖股票的最佳时间 IV

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 k transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Example 1:

Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
             Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

123. Best Time to Buy and Sell Stock III 这题是最多能交易2次，而这题是最多k次。

要用动态规划Dynamic programming来解，需要两个递推公式来分别更新两个变量local和global。定义local[i][j]为在到达第i天时最多可进行j次交易并且最后一次交易在最后一天卖出的最大利润，此为局部最优。然后我们定义global[i][j]为在到达第i天时最多可进行j次交易的最大利润，此为全局最优。它们的递推式为：

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])

Java:

public int maxProfit(int k, int[] prices) {
        int len = prices.length;
        if (k >= len / 2) return quickSolve(prices);

        int[][] t = new int[k + 1][len];
        for (int i = 1; i <= k; i++) {
            int tmpMax =  -prices[0];
            for (int j = 1; j < len; j++) {
                t[i][j] = Math.max(t[i][j - 1], prices[j] + tmpMax);
                tmpMax =  Math.max(tmpMax, t[i - 1][j - 1] - prices[j]);
            }
        }
        return t[k][len - 1];
    }

    private int quickSolve(int[] prices) {
        int len = prices.length, profit = 0;
        for (int i = 1; i < len; i++)
            // as long as there is a price gap, we gain a profit.
            if (prices[i] > prices[i - 1]) profit += prices[i] - prices[i - 1];
        return profit;
    }

Python:

class Solution(object):
    # @return an integer as the maximum profit
    def maxProfit(self, k, prices):
        if k >= len(prices) / 2:
            return self.maxAtMostNPairsProfit(prices)

        return self.maxAtMostKPairsProfit(prices, k)

    def maxAtMostNPairsProfit(self, prices):
        profit = 0
        for i in xrange(len(prices) - 1):
            profit += max(0, prices[i + 1] - prices[i])
        return profit

    def maxAtMostKPairsProfit(self, prices, k):
        max_buy = [float("-inf") for _ in xrange(k + 1)]
        max_sell = [0 for _ in xrange(k + 1)]

        for i in xrange(len(prices)):
            for j in xrange(1, min(k, i/2+1) + 1):
                max_buy[j] = max(max_buy[j], max_sell[j-1] - prices[i])
                max_sell[j] = max(max_sell[j], max_buy[j] + prices[i])

        return max_sell[k]　　　　

C++:

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

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