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## 题目描述

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.

## 代码

### Approach #1 Two Array

public class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length <= 1) return 0;
int[] profitFront = new int[prices.length];
profitFront = 0;
int leftMin = prices;
for (int i = 1; i < prices.length; i++) {
profitFront[i] = Math.max(profitFront[i - 1], prices[i] - leftMin);
leftMin = Math.min(leftMin, prices[i]);
}
int[] profitBack = new int[prices.length];
profitBack[prices.length - 1] = 0;
int rightMax = prices[prices.length - 1];
for (int i = prices.length - 2; i >= 0; i--) {
proftiBack[i] = Math.max(profitBack[i + 1], rightMax - prices[i]);
rightMax = Math.max(rightMax, prices[i]);
}
int profit = 0;
for (int i = 0; i < prices.length; i++) {
profit = Math.max(profit, profitFront[i] + profitBack[i]);
}
return profit;
}
}

### Approach #2

sell1 means we decide to sell the stock, after selling it we have price[i] money and we have to give back the money we owed, so we have price[i] - |buy1| = prices[i ] + buy1, we want to make this max.
buy2 means we want to buy another stock, we already have sell1 money, so after buying stock2 we have buy2 = sell1 - price[i] money left, we want more money left, so we make it max
sell2 means we want to sell stock2, we can have price[i] money after selling it, and we have buy2 money left before, so sell2 = buy2 + prices[i], we make this max.
So sell2 is the most money we can have.
class Solution {
public int maxProfit(int[] prices) {
int sell1 = 0, sell2 = 0;