Best Time to Buy and Sell Stock III
2013-06-05 10:54
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题目:
Say you have an array for which the ith element is the price of a given stock on dayi.
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).
代码如下:
int maxProfit(vector<int> &prices) {
int n=prices.size();
if(n<=1)return 0;
vector<int> left(n,0);
vector<int> right(n,0);
int minl=prices[0];
for(int i=1;i<n;i++)
{
minl=std::min(minl,prices[i]);
left[i]=std::max(left[i-1],prices[i]-minl);
}
int maxr=prices[n-1];
for(int i=n-2;i>=0;i--)
{
maxr=std::max(maxr,prices[i]);
right[i]=std::max(right[i+1],maxr-prices[i]);
}
int sum=0;
for (int i=0;i<n;i++)
sum=std::max(sum,left[i]+right[i]);
return sum;
}
Say you have an array for which the ith element is the price of a given stock on dayi.
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).
代码如下:
int maxProfit(vector<int> &prices) {
int n=prices.size();
if(n<=1)return 0;
vector<int> left(n,0);
vector<int> right(n,0);
int minl=prices[0];
for(int i=1;i<n;i++)
{
minl=std::min(minl,prices[i]);
left[i]=std::max(left[i-1],prices[i]-minl);
}
int maxr=prices[n-1];
for(int i=n-2;i>=0;i--)
{
maxr=std::max(maxr,prices[i]);
right[i]=std::max(right[i+1],maxr-prices[i]);
}
int sum=0;
for (int i=0;i<n;i++)
sum=std::max(sum,left[i]+right[i]);
return sum;
}
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