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.
题目大意：与之前的题目类似，不过这次有次数限制。最多两次买卖，问最大获利。
思路：如果将数组拆成两半，那么每一半都可以用Best Time to Buy and Sell Stock I 题目中的方式获取到最大获利，前后加起来，就是最大获利。用left[i]表示从0到i的最大获利，遍历一遍可求得。用right[i]表示从i到length-1的最大获利，从右向左遍历一遍可获得。从左往右遍历时，一边找最小price一边更新max获利；从右往左遍历时，一边找最大price，一边更新当前max获利。两遍遍历，时间复杂度O(n)，讨论区还有一遍遍历的解法，有兴趣可以去看看。
public static int maxProfit(int[] prices) {
if (prices == null || prices.length <= 1) {
return 0;
}
int[] left = new int[prices.length];
int[] right = new int[prices.length];
int min = prices[0];
for (int i = 1; i < prices.length; i++) {
left[i] = Math.max(prices[i] - min, left[i - 1]);
min = Math.min(min, prices[i]);
}
int max = prices[prices.length - 1];
int res = 0;
for (int i = prices.length - 2; i >= 0; i--) {
right[i] = Math.max(max - prices[i], right[i + 1]);
max = Math.max(max, prices[i]);
res = Math.max(left[i] + right[i], res);
}
return res;
}