Best Time to Buy and Sell Stock III

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 (ie, you must sell the stock before you buy again).

Code:

Method1:

class Solution {
public:
int maxProfit(vector<int> &prices) {
int oneTry=0,oneHis=0,twoTry=0,twoHis=0; //Try: attempt to buy; His: the max-profit record; One/Two: one/two transaction(s)
int n = prices.size() - 1;
for (int i=0;i<n;i++){
int diff = prices[i+1] - prices[i];
if(i>0){
twoTry = max(twoTry, oneHis) + diff;
twoHis = max(twoTry,twoHis); // the max-profit by totally two transactions
}
oneTry = max(oneTry, 0) + diff;
oneHis = max(oneTry,oneHis); // the max-profit by totally one transaction
}
return max(oneHis,twoHis);
}
};

Method2:

class Solution {
public:
int maxProfit(vector<int> &prices) {
// null check
int len = prices.size();
if (len==0) return 0;

vector<int> historyProfit;
vector<int> futureProfit;
historyProfit.assign(len,0);
futureProfit.assign(len,0);
int valley = prices[0];
int peak = prices[len-1];
int maxProfit = 0;

// forward, calculate max profit until this time
for (int i = 0; i<len; ++i)
{
valley = min(valley,prices[i]);
if(i>0)
{
historyProfit[i]=max(historyProfit[i-1],prices[i]-valley);
}
}

// backward, calculate max profit from now, and the sum with history
for (int i = len-1; i>=0; --i)
{
peak = max(peak, prices[i]);
if (i<len-1)
{
futureProfit[i]=max(futureProfit[i+1],peak-prices[i]);
}
maxProfit = max(maxProfit,historyProfit[i]+futureProfit[i]);
}
return maxProfit;
}
};