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).
解题思路：
既然是两次交易的话，分为左右两个区间即可。先按照一次交易的思路算出交易的最大值，存一个数组里，然后从后往前遍历，找到符合条件的和最大的两个，JAVA实现如下：
public int maxProfit(int[] prices) {
if (prices.length == 0)
return 0;
int[] oneProfit = new int[prices.length];
int buy_price = prices[0], profit = 0;
for (int i = 1; i < prices.length; i++) {
buy_price = Math.min(buy_price, prices[i]);
profit = Math.max(profit, prices[i] - buy_price);
oneProfit[i] = profit;
}
int res = oneProfit[prices.length - 1];
int sell_price = prices[prices.length - 1];
profit = 0;
for (int i = prices.length - 1; i >= 1; i--) {
sell_price = Math.max(sell_price, prices[i]);
profit = Math.max(profit, sell_price - prices[i]);
res = Math.max(res, profit + oneProfit[i - 1]);
}
return res;
}
Java for LeetCode 123 Best Time to Buy and Sell Stock III
原文：http://www.cnblogs.com/tonyluis/p/4530984.html