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 (i.e., you must sell the stock before you buy again).
Example 1:
Input: [3,3,5,0,0,3,1,4] Output: 6 Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:
Input: [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
考察:dp;
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.empty())
return 0;
vector<int> sum(prices.size(), 0);
int buy = INT_MAX, res = 0;
for (int i = 0; i < prices.size(); i ++) {
buy = min(buy, prices[i]);
res = max(res, prices[i]-buy);
sum[i] += res;
}
buy = INT_MIN, res = 0;
for (int i = prices.size()-1; i >= 0; i --) {
buy = max(buy, prices[i]);
res = max(res, buy-prices[i]);
sum[i] += res;
}
return *(max_element(sum.begin(), sum.end()));
}
};
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.empty())
return 0;
int res1 = 0, res2 = 0;
int buy1 = INT_MIN, buy2 = INT_MIN;
for (int i = 0; i < prices.size(); i ++) {
buy1 = max(buy1, -prices[i]); // 第一次买完
res1 = max(res1, buy1+prices[i]); // 第一次卖完
buy2 = max(buy2, res1-prices[i]); // 第二次买完
res2 = max(res2, buy2+prices[i]); // 第二次卖完
}
return res2;
}
};