文章标题

题目描述

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).

定义两个数组，before从前向后遍历，第i个位置代表前i天买卖一次的最大收益

after从后向前遍历，第i个位置代表i天之后买卖一次的最大收益

最后从前向后遍历一遍，找出before[i]和after[i]和的最大值即可

class Solution {
public:
int maxProfit(vector<int> &prices)
{
if(prices.size()<=1)
return 0;
int sz=prices.size();
vector<int> before(sz,0);
int min_val=prices[0];
int max_mon=0;
for(int i=1;i<sz;++i)
{
if(prices[i]-min_val>max_mon)
{
max_mon=prices[i]-min_val;
}
if(prices[i]<min_val)
min_val=prices[i];
before[i]=max(prices[i]-min_val,before[i-1]);
}
int max_val=prices[sz-1];
max_mon=0;
vector<int> after(sz,0);
for(int i=sz-2;i>=0;--i)
{
if(max_val-prices[i]>max_mon)
max_mon=max_val-prices[i];
if(max_val<prices[i])
max_val=prices[i];
after[i]=max(after[i+1],max_val-prices[i]);
}
int maxsum=0;
for(int i=0;i<sz;++i)
{
if(before[i]+after[i]>maxsum)
maxsum=before[i]+after[i];
}
return maxsum;
}

};