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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.（股票交易，最多两次）
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
class Solution {
public:
int maxProfit(vector<int> &prices) {
if(prices.size()==0)return 0;
int n=prices.size();
vector<int> left(n);
vector<int> right(n);
int min=prices[0];
int max=prices[n-1];
int res=0;
for(int i=1;i<n;i++)
{
min=min<prices[i]?min:prices[i];
left[i]=left[i-1]>(prices[i]-min)?left[i-1]:(prices[i]-min);
}
for(int j=n-2;j>=0;j--)
{
max=max>prices[j]?max:prices[j];
right[j]=right[j+1]>(max-prices[j])?right[j+1]:(max-prices[j]);
}
for(int i=0;i<n;i++)
{
res=res>(left[i]+right[i])?res:(left[i]+right[i]);
}
return res;
}
};
最多两次股票交易-Best Time to Buy and Sell Stock III
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原文地址：http://www.cnblogs.com/Vae1990Silence/p/4830586.html
