123. Best Time to Buy and Sell Stock III（js）

123. Best Time to Buy and Sell Stock III

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.
题意：最多两次买卖，求最大收益
代码如下：

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    
    let buy1=-Infinity,buy2=-Infinity
    let sell1=0,sell2=0
    prices.forEach(item=>{
        sell2=Math.max(sell2,buy2+item);
        buy2=Math.max(buy2,sell1-item);
        sell1=Math.max(sell1,buy1+item);
        buy1=Math.max(buy1,-item)
    })
    return sell2;
}