[LeetCode] Best Time to Buy and Sell Stock III 买股票的最佳时间之三

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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 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.

这道是买股票的最佳时间系列问题中最难最复杂的一道，前面两道 Best Time to Buy and Sell Stock 和 Best Time to Buy and Sell Stock II 的思路都非常的简洁明了，算法也很简单。而这道是要求最多交易两次，找到最大利润，还是需要用动态规划Dynamic Programming来解，而这里我们需要两个递推公式来分别更新两个变量local和global，参见网友Code Ganker的博客，我们其实可以求至少k次交易的最大利润，找到通解后可以设定 k = 2，即为本题的解答。我们定义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])

其中局部最优值是比较前一天并少交易一次的全局最优加上大于0的差值，和前一天的局部最优加上差值中取较大值，而全局最优比较局部最优和前一天的全局最优，代码如下：

解法一：

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

下面这种解法用一维数组来代替二维数组，可以极大的节省了空间，由于覆盖的顺序关系，我们需要j从2到1，这样可以取到正确的g[j-1]值，而非已经被覆盖过的值，参见代码如下：

解法二：

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

我们如果假设prices数组为1, 3, 2, 9, 那么我们来看每次更新时local 和 global 的值：

第一天两次交易：第一天一次交易：

local: 0 0 0 local: 0 0 0

global:0 0 0 global:0 0 0

第二天两次交易：第二天一次交易：

local: 0 0 2 local: 0 2 2

global:0 0 2 global:0 2 2

第三天两次交易：第三天一次交易：

local: 0 2 2 local: 0 1 2

global:0 2 2 global:0 2 2

第四天两次交易：第四天一次交易：

local: 0 1 9 local: 0 8 9

global:0 2 9 global:0 8 9

在网友@loveahnee的提醒下，发现了其实上述的递推公式关于local[i][j]的可以稍稍化简一下，我们之前定义的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，然后网友@fgvlty解释了一下第 i 天卖第 j 支股票的话，一定是下面的一种：

1. 今天刚买的
那么 Local(i, j) = Global(i-1, j-1)
相当于啥都没干

2. 昨天买的
那么 Local(i, j) = Global(i-1, j-1) + diff
等于Global(i-1, j-1) 中的交易，加上今天干的那一票

3. 更早之前买的
那么 Local(i, j) = Local(i-1, j) + diff
昨天别卖了，留到今天卖

但其实第一种情况是不需要考虑的，因为当天买当天卖不会增加利润，完全是重复操作，这种情况可以归纳在global[i-1][j-1]中，所以我们就不需要max(0, diff)了，那么由于两项都加上了diff，所以我们可以把diff抽到max的外面，所以更新后的递推公式为：

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

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

类似题目：

Best Time to Buy and Sell Stock with Cooldown

Best Time to Buy and Sell Stock IV

Best Time to Buy and Sell Stock II

Best Time to Buy and Sell Stock

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