Best Time to Buy and Sell Stock I,II,III [leetcode]

Best Time to Buy and Sell Stock I

只能作一次操作时：维护preMin记录之前出现的最小值

代码如下：

int maxProfit(vector<int> &prices) {
if (prices.size() == 0) return 0;
int profit = 0;
int preMin = prices[0];
for (int i = 1; i < prices.size(); i++)
{
if (prices[i] < preMin) preMin = prices[i];
profit = max(profit, prices[i] - preMin);
}
return profit;
}

Best Time to Buy and Sell Stock II

能无限次操作时：当==>当前价格 > 买入价格的时候，卖出

代码如下：

int maxProfit(vector<int> &prices) {
if (prices.size() == 0) return 0;
int profit = 0;
int buy = prices[0];
for (int i = 1; i < prices.size(); i++)
{
if (buy < prices[i])
profit += prices[i] - buy;
buy = prices[i];
}
return profit;
}

Best Time to Buy and Sell Stock III

只能操作两次时：

两次操作不能重叠，可以分成两部分：0...i的最大利润fst 和i...n-1的最大利润snd

代码如下：

int maxProfit(vector<int> &prices) {
if (prices.size() == 0) return 0;
int size = prices.size();
vector<int> fst(size);
vector<int> snd(size);

int preMin = prices[0];
for (int i = 1; i < size; i++)
{
if (preMin > prices[i]) preMin = prices[i];
fst[i] = max(fst[i - 1], prices[i] - preMin);
}
int profit = fst[size - 1];
int postMax = prices[size - 1];
for (int i = size - 2; i >= 0; i--)
{
if (postMax < prices[i]) postMax = prices[i];
snd[i] = max(snd[i + 1], postMax - prices[i]);
//update profit
profit = max(profit, snd[i] + fst[i]);
}
return profit;
}