123.Best-Time-to-Buy-and-Sell-Stock-III.md

123. Best Time to Buy and Sell Stock III

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题目地址

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/

https://www.lintcode.com/problem/best-time-to-buy-and-sell-stock-iii

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] = 0;
    int leftMin = prices[0];
    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;
    int buy1 = Integer.MIN_VALUE, buy2 = Integer.MIN_VALUE;
    for (int i = 0; i < prices.length; i++) {
      buy1 = Math.max(buy1, -prices[i]);
      sell1 = Math.max(sell1, buy1 + prices[i]);
      
      buy2 = Math.max(buy2, sell1 - prices[i]);
      sell2 = Math.max(sell2, buy2 + prices[i]);
    }
    
    return sell2;
  }
}