# 123. Best Time to Buy and Sell Stock III

## 刷题内容

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.


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class Solution {
public:
int maxProfit(vector<int>& prices) {
int len = prices.size();
if(!len)
return 0;
int arr[len + 1];
memset(arr,0,sizeof(arr));
int max1 = prices[len - 1],price = 0;
for(int j = len - 2;j >= 0;--j)
{
if(prices[j] > max1)
max1 = prices[j];
arr[j] = max(arr[j + 1],max1 - prices[j]);
}
int ret = 0;
for(int i = 0;i < len;++i)
for(int j = i + 1;j < len;++j)
if(prices[j] > prices[i])
ret = max(ret,prices[j] - prices[i] + arr[j + 1]);
return ret;
}
};


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`cpp class Solution { public: int maxProfit(vector& prices) { int len = prices.size(); if(!len) return 0; int arr[len + 1]; memset(arr,0,sizeof(arr)); int max1 = prices[len - 1],price = 0; for(int j = len - 2;j >= 0;--j) { if(prices[j] > max1) max1 = prices[j]; arr[j] = max(arr[j + 1],max1 - prices[j]); } int ret = 0; int min1 = prices; for(int i = 0;i < len;++i) { if(prices[i] < min1) min1 = prices[i]; ret = max(ret,prices[i] - min1 + arr[i + 1]); } return ret; } };