算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Best Time to Buy and Sell Stock III


### 描述

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 two transactions.

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


### 分析

设状态`f(i)`,表示区间$$[0,i](0 \leq i \leq n-1)$$的最大利润,状态`g(i)`,表示区间$$[i, n-1](0 \leq i \leq n-1)$$的最大利润,则最终答案为$$\max\left\{f(i)+g(i)\right\},0 \leq i \leq n-1$$。

允许在一天内买进又卖出,相当于不交易,因为题目的规定是最多两次,而不是一定要两次。

将原数组变成差分数组,本题也可以看做是最大`m`子段和,`m=2`,参考代码:


### 代码

{% if book.java %}
{% codesnippet "./code/best-time-to-buy-and-sell-stock-iii."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Best Time to Buy and Sell Stock III
// 时间复杂度O(n),空间复杂度O(n)
class Solution {
public:
int maxProfit(vector& prices) {
if (prices.size() < 2) return 0;

const int n = prices.size();
vector f(n, 0);
vector g(n, 0);

for (int i = 1, valley = prices[0]; i < n; ++i) {
valley = min(valley, prices[i]);
f[i] = max(f[i - 1], prices[i] - valley);
}

for (int i = n - 2, peak = prices[n - 1]; i >= 0; --i) {
peak = max(peak, prices[i]);
g[i] = max(g[i], peak - prices[i]);
}

int max_profit = 0;
for (int i = 0; i < n; ++i)
max_profit = max(max_profit, f[i] + g[i]);

return max_profit;
}
};
```
{% endif%}


### 相关题目

* [Best Time to Buy and Sell Stock](../greedy/best-time-to-buy-and-sell-stock.md)
* [Best Time to Buy and Sell Stock II](../greedy/best-time-to-buy-and-sell-stock-ii.md)