【LeetCode】解题188：Best Time to Buy and Sell Stock IV（动态规划）

Problem 188: Best Time to Buy and Sell Stock IV [Hard]

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 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.

来源：LeetCode

解题思路

本题需要使用动态规划思想。与题解123：Best Time to Buy and Sell Stock III同属于最佳时机买卖股票问题。

解题思路选用[题解123]中的方法III，123题中是最多可以交易2次，复杂度为$O(n)$时间+$O(4)$空间，本题中最多可以交易k次，所以复杂度为$O(n)$时间+$O(2k)$空间。
在一次循环中，可以使用k组{minCost_j, maxP_j}，先计算第一轮的{minCost₁, maxP₁}，再计算第二轮的{minCost₂, maxP₂}，…，最后计算第k轮的{minCost_k, maxP_k}。

更新公式为：
$$cpc{p}_0=\left[\begin{array}{c}prices[0]\\ 0\\ prices[0]\\ 0\\ ...\\ prices[0]\\ 0\end{array}\right]$$
c p c p = [ { min ⁡ { c p c p [ 0 ] , p r i c e s [ i ] } max ⁡ { c p c p [ 1 ] , p r i c e s [ i ] − c p c p [ 0 ] } . . . min ⁡ { c p c p [ 2 ∗ j ] , p r i c e s [ i ] − c p c p [ 2 ∗ j − 1 ] } max ⁡ { c p c p [ 2 ∗ j + 1 ] , p r i c e s [ i ] − c p c p [ 2 ∗ j ] } . . . min ⁡ { c p c p [ 2 ∗ k − 2 ] , p r i c e s [ i ] − c p c p [ 2 ∗ k − 3 ] } max ⁡ { c p c p [ 2 ∗ k − 1 ] , p r i c e s [ i ] − c p c p [ 2 ∗ k − 2 ] } ] , i ∈ [ 0 , N − 1 ] , j ∈ [ 0 , k − 1 ] cpcp = \begin{bmatrix} \{\min\{cpcp[0], prices[i]\} \\ \max\{cpcp[1], prices[i]-cpcp[0]\} \\ ... \\ \min\{cpcp[2*j], prices[i] - cpcp[2*j-1]\} \\ \max\{cpcp[2*j+1], prices[i] - cpcp[2*j]\} \\ ... \\ \min\{cpcp[2*k-2], prices[i] - cpcp[2*k-3]\} \\ \max\{cpcp[2*k-1], prices[i] - cpcp[2*k-2]\} \end{bmatrix}, i\in[0, N-1], j\in[0, k-1] cpcp=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​{min{cpcp[0],prices[i]}max{cpcp[1],prices[i]−cpcp[0]}...min{cpcp[2∗j],prices[i]−cpcp[2∗j−1]}max{cpcp[2∗j+1],prices[i]−cpcp[2∗j]}...min{cpcp[2∗k−2],prices[i]−cpcp[2∗k−3]}max{cpcp[2∗k−1],prices[i]−cpcp[2∗k−2]}​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​,i∈[0,N−1],j∈[0,k−1]
其中$N$为总天数，$k$为最大交易次数。
c p c p = { m i n C o s t 1 , m a x P 1 , . . . , m i n C o s t j + 1 , m a x P j + 1 , . . . , m i n C o s t k , m a x P k } cpcp = \{minCost_1, maxP_1, ...,minCost_{j+1}, maxP_{j+1},...,minCost_k, maxP_k\} cpcp={minCost1​,maxP1​,...,minCostj+1​,maxPj+1​,...,minCostk​,maxPk​}，更新过程相当于
c p c p = [ m i n C o s t 1 = min ⁡ { m i n C o s t 1 , p r i c e s [ i ] } m a x P 1 = max ⁡ { m a x P 1 , p r i c e s [ i ] − m i n C o s t 1 } . . . m i n C o s t j + 1 = min ⁡ { m i n C o s t j + 1 , p r i c e s [ i ] − m a x P j } m a x P j + 1 = max ⁡ { m a x P j + 1 , p r i c e s [ i ] − m i n C o s t j + 1 } . . . m i n C o s t k = min ⁡ { m i n C o s t k , p r i c e s [ i ] − m a x P k − 1 } m a x P k = max ⁡ { m a x P k , p r i c e s [ i ] − m i n C o s t k } ] cpcp = \begin{bmatrix} minCost_1 = \min\{minCost_1,prices[i]\} \\ maxP_1 = \max\{maxP_1,prices[i]-minCost_1\} \\ ... \\ minCost_{j+1} = \min\{minCost_{j+1},prices[i]-maxP_j\} \\ maxP_{j+1} = \max\{maxP_{j+1}, prices[i]-minCost_{j+1}\} \\ ... \\ minCost_k = \min\{minCost_k,prices[i]-maxP_{k-1}\} \\ maxP_k = \max\{maxP_k,prices[i]-minCost_k\} \end{bmatrix} cpcp=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​minCost1​=min{minCost1​,prices[i]}maxP1​=max{maxP1​,prices[i]−minCost1​}...minCostj+1​=min{minCostj+1​,prices[i]−maxPj​}maxPj+1​=max{maxPj+1​,prices[i]−minCostj+1​}...minCostk​=min{minCostk​,prices[i]−maxPk−1​}maxPk​=max{maxPk​,prices[i]−minCostk​}​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​
该方法只需要$O(n)$时间，$O(2k)$空间。

注意：当k非常大时，会出现超出内存限制的问题，此时需要将$k>\frac{N}{2}$的情况分开考虑。由于一次交易至少需要两天，所以当$k>\frac{N}{2}$时，实际上相当于交易次数不受限制，也就是题解122：Best Time to Buy and Sell Stock II中的题意，可以复用该题的代码。

运行结果：

要点：动态规划

Solution (Java)

O(n)时间+O(2k)空间

class Solution {
    public int maxProfit(int k, int[] prices) {
        int N = prices.length;
        if(k < 1 || N < 2) return 0;
        if(k > N/2) return nolimit(prices);
        int[] cpcp = new int[2*k];
        for(int j = 0; j < k; j++){
            cpcp[2*j] = prices[0];
            cpcp[2*j+1] = 0;
        }
        for(int i = 1; i < N; i++){
            cpcp[0] = Math.min(cpcp[0], prices[i]);
            cpcp[1] = Math.max(cpcp[1], prices[i] - cpcp[0]);
            for(int j = 1; j < k; j++){
                cpcp[2*j] = Math.min(cpcp[2*j], prices[i] - cpcp[2*j-1]);
                cpcp[2*j+1] = Math.max(cpcp[2*j+1], prices[i] - cpcp[2*j]);
            }
        }
        return cpcp[2*k-1];
    }
    private int nolimit(int[] prices){
        int maxP = 0;
        for(int i = 1; i < prices.length; i++){
            maxP += Math.max(0, prices[i] - prices[i-1]);
        }
        return maxP;
    }
}

修改过程

• 样例输入k=1000000000时，出现超出内存限制问题，需要将其考虑为无限次交易的情况，因此添加函数nolimit()。