标签：most target 最大的 targe com end bsp csdn lan
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).
开始的时候思路很朴素：以为与leetcode-121-Best Time to Buy and Sell Stock差不多
就是选取一个分割点j，在j左边与右边分别进行计算最大值，然后统计最大值之和。时间复杂度为O(n2).没有通过
最后一个数据量特别大的测试用例。
int maxProfitcore(vector<int>& prices,int start,int end) {//从start 到 end期间 最大profit if (start == end|| prices.size()==0) return 0; int tempProfit = 0, maxProfit = 0; for (int i = start+1; i <=end; i++) { tempProfit += prices[i] - prices[i - 1]; tempProfit = max(0, tempProfit); maxProfit = max(maxProfit, tempProfit); } return maxProfit; }//超时 int maxProfit3(vector<int>& prices) { int profit = 0; for (int i = 1; i < prices.size();i++) { int pro1 = maxProfitcore(prices, 0, i); int pro2 = maxProfitcore(prices, i, prices.size()-1); profit = max(profit,pro1+pro2); } return profit; }
只需扫描一趟就行，复杂度为O(n).
如下：
int maxProfitnew(vector<int>& prices) { int size = prices.size(); if (size == 0 || size == 1) return 0; vector<int>profit(size,0); vector<int>profit1(size); int local_min = prices[0]; int local_max = prices[size - 1]; int j = size - 2; int result = 0; profit[0] = 0; profit1[size - 1] = 0; for (int i = 1; i < size + 1 && j >= 0; i++, j--)//扫描一遍，分别计算左边和右边最大的profit { //记录i左边最大profit 利用i-1的信息 只需比较i-1时候的profit 与现在price-local_min即可 //如果prices[i] - local_min >profit[i - 1]的话，说明需要更新最大的profit 同理 求右边最大profit也是一个道理 profit[i] = max(profit[i - 1], prices[i] - local_min); local_min = min(local_min, prices[i]); profit1[j] = max(profit1[j + 1], local_max - prices[j]);//记录j右边最大profit local_max = max(local_max, prices[j]); } for (int i = 1; i < size; i++)//统计左右两边最大profit之和 取最大 { result = max(result, profit[i] + profit1[i]); } return result; }
参考：
http://blog.csdn.net/fightforyourdream/article/details/14503469
[leetcode-123-Best Time to Buy and Sell Stock III]
标签：most target 最大的 targe com end bsp csdn lan
原文地址：http://www.cnblogs.com/hellowooorld/p/6492413.html
