# 123. Best Time to Buy and Sell Stock III

## Description

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 2:

``````Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
``````

Tags: Math, String

## 题解

### 思路1

• 遍历一遍数组，求`[0,i−1][0,i−1`]区间的最大利润`f(i)`，具体做法是找当前最低价格`low`，判断是要以`low`买入当天卖出，还是不动
• 从后往前遍历，求`[i,n−1][i,n−1]`区间的最大利润`g(i)`，具体做法是找当前最高价格`high`，判断是要当天买入以`high`卖出，还是不动
• 遍历，求最大利润`max(f(i)+g(i))`
``````func maxProfit(prices []int) int {
t1_b, t1_s, t2_b, t2_s := math.MinInt32, 0, math.MinInt32, 0

for _, v := range prices {
t1_b = max(t1_b, 0-v)
t1_s = max(t1_s, t1_b+v)
t2_b = max(t2_b, t1_s-v)
t2_s = max(t2_s, t2_b+v)
}
return t2_s
}
func max(a, b int) int {
if a < b {
return b
}
return a
}
``````