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).
第一次尝试，超时时间复杂度O(n^2)。

class Solution {
public:
int max(vector<int>& prices, int s, int e)
{
if (s >= e) return 0;
int ret = 0;
int min = prices[s];
for (int i = s + 1; i <= e; i++)
{
if (prices[i] < min)
min = prices[i];
else if (prices[i] - min > ret)
{
ret = prices[i] - min;
}
}
return ret;
}
int maxProfit(vector<int> &prices)
{
if (prices.size() <= 1) return 0;
int ret = 0;
for (int i = 0; i < prices.size(); i++)
{
int cur = max(prices, 0, i) + max(prices, i, prices.size() - 1);
ret = ret > cur ? ret : cur;
}
return ret;
}
};

第二次通过,利用动态规划时间复杂度O(n),逆序遍历与顺序不同的是，逆序是记录最大值并于当前值比较，顺序是记录最小值，并与当前值比较

class Solution {
public:
int maxProfit(vector<int> &prices)
{
if (prices.size() <= 1) return 0;

int N = prices.size();
vector<int> dp;
int min = prices[0];

dp.push_back(0);
for (int i = 1; i < N; i++)
{
min = min > prices[i] ? prices[i] : min;
int tmp = prices[i] - min;
dp.push_back(tmp > dp[i-1] ? tmp : dp[i-1]);
}
int max = prices[N - 1];
int maxProf = 0;
int ret = 0;
for (int i = N-2; i >= 0; i--)
{
max = max > prices[i] ? max : prices[i];
maxProf = maxProf > max - prices[i] ? maxProf : max - prices[i];
ret = ret > maxProf + dp[i] ? ret : maxProf + dp[i];
}
return ret;
}
};