LeetCode Best Time to Buy and Sell Stock III

Description:

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.
Solution:

这道题目的变化是，至多可以买卖两次。

可以利用和1一样的想法，只不过1只需要从后往前遍历一次，或者从前往后遍历一次，这里就直接结合。考虑的想法是，既然买卖两次，可以以某个点为断点，前部分找一次最大，后部分找一次最大；然后遍历所有的点，将每个点前后两次最大利润加起来，取这个利润和的最大值。

import java.util.*;

public class Solution {
public int maxProfit(int[] prices) {
int n = prices.length;
if (n == 0)
return 0;

int after[] = new int
;
int before[] = new int
;

int maxPrice = prices[n - 1], max = 0;
for (int i = n - 1; i >= 0; i--) {
maxPrice = Math.max(maxPrice, prices[i]);
max = Math.max(max, maxPrice - prices[i]);
after[i] = max;
}

int minPrice = prices[0];
max = 0;
for (int i = 0; i < n; i++) {
minPrice = Math.min(minPrice, prices[i]);
max = Math.max(max, prices[i] - minPrice);
before[i] = max;
}

max = 0;
for (int i = 0; i < n; i++)
max = Math.max(max, after[i] + before[i]);
return max;
}
}