123. Best Time to Buy and Sell Stock III
难度:Hard
刷题内容
原题连接
- https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/
内容描述
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.
˼·1 **- ʱ�临�Ӷ�: O(n^2)*- �ռ临�Ӷ�: O(n)***
��֮ǰ�IJ�ͬ�������Ҫ�������Σ内容描述�ü��仯�ķ�����¼�µ�i���֮�������ֵ������ǵڶ���ÿ�콻��֮������ֵ������ֻҪ�������飬�����һ�콻�����ֵ����һ内容描述ö����ÿ�ν��ף内容描述�ֵ��
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;
}
};
˼·2 **- ʱ�临�Ӷ�: O(n)*- �ռ临�Ӷ�: O(n)***
�ڶ��ַ������Ƕ�˼·1�������Ż内容描述һ�ν���ʱ����һ����Сֵmin1�������˵�i��֮ǰ����Сֵ��֮内容描述飬ÿ�θ���min1�����prices[i] - min1�����ֵ����
```cpp
class Solution {
public:
int maxProfit(vector