LeetCode123 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. （Hard）

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

分析：

最多购买两次股票且不能同时拥有，所以买卖两次肯定分布在前后两个不同的区间。

设dp(i) = 区间[0,1,2...i]的最大利润 + 区间[i,i+1,....n-1]的最大利润,那么本题的最大利润 = max{dp[0],dp[1],dp[2],...,dp[n-1]}.

而对于式子中两个区间内分别只能有一次买卖，这是Best Time to Buy and Sell Stock I的问题。

代码：

1 class Solution {
2 public:
3     int maxProfit(vector<int> &prices) {
4         int len = prices.size();
5         if(len <= 1) {
6             return 0;
7         }
8         int maxFromHead[len];
9         maxFromHead[0] = 0;
10         int minprice = prices[0], maxprofit = 0;
11         for(int i = 1; i < len; i++) {
12             minprice = min(prices[i-1], minprice);
13             if(maxprofit < prices[i] - minprice)
14                 maxprofit = prices[i] - minprice;
15             maxFromHead[i] = maxprofit;
16         }
17         int maxprice = prices[len - 1];
18         int res = maxFromHead[len-1];
19         maxprofit = 0;
20         for(int i = len-2; i >=0; i--) {
21             maxprice = max(maxprice, prices[i+1]);
22             if(maxprofit < maxprice - prices[i])
23                 maxprofit = maxprice - prices[i];
24             if(res < maxFromHead[i] + maxprofit)
25                 res = maxFromHead[i] + maxprofit;
26         }
27         return res;
28     }
29 };