Best Time to Buy and Sell Stock III Java

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 (ie, you must sell the stock before you buy again).

The extension of of "Best Time to Buy and Sell Stock"
    Solve by DP Algo again
    this time we may complete at most two transaction instead
    of at most one transaction in previous problem of Best Time
    to Buy and Sell Stock. Think of in Big picture of problem in
    solve question by complete at most k transaction, k = 2.

    Best Time to Buy and Sell Stock III
    i: day i
    j: transaction time j that 1<=j<=k
    then we have DP formula:
    diff: Profit difference between previous and current day since at most we can do twice
            transaction a day
    Optimize Local: local[j]=Max(global[j-1]+max(diff,0), local[j]+diff) in day i-1
    Optimize Global: global[j]=Max(local[j],global[j])  in day i-1

    Time: O(N*k) k=2 in here => O(N)
    Space: O(k)



public class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length==0) return 0;
        int k=2; // k is at most 2 times
        int local[]=new int[k+1];
        int global[]=new int[k+1];
        for(int i=0;i<prices.length-1;i++){ //days start from i-1
         int diff=prices[i+1]-prices[i];
            for(int j=k;j>=1;j--){
                //optimize local value base on previous global[j-1] (first) transaction
                local[j]=Math.max(global[j-1]+Math.max(diff,0),local[j]+diff);
                global[j]=Math.max(local[j],global[j]);
            }
        }
        return global[k];
    }
}


最佳的时间买卖股票III问题可以使用贪心算法来解决。该问题要求在最多进行两次交易的情况下,获取最大的利润。 贪心法的思路是通过在每一天进行买入和卖出操作来获取最大利润。我们可以定义四个变量:buy1、sell1、buy2和sell2,分别表示第一次买入、第一次卖出、第二次买入和第二次卖出的利润。 我们首先将buy1和buy2初始化为正无穷大,sell1和sell2初始化为0。然后遍历股票价格列表,更新这些变量的值。 对于每一天的股票价格,我们可以尝试更新第一次买入的价格和利润。如果当前股票价格比buy1小,我们更新buy1为当前价格。否则,我们计算当前价格与buy1的差值,如果大于sell1,则将sell1更新为该差值。 接下来,我们尝试更新第二次买入的价格和利润。如果当前股票价格减去sell1比buy2小,我们更新buy2为当前价格减去sell1。否则,我们计算当前价格减去sell1的差值,如果大于sell2,则将sell2更新为该差值。 最后,我们返回sell2作为最大利润。 下面是使用贪心算法解决最佳的时间买卖股票III问题的代码示例(假设prices是股票价格的列表): ```python def maxProfit(prices): buy1 = float('inf') buy2 = float('inf') sell1 = 0 sell2 = 0 for price in prices: buy1 = min(buy1, price) sell1 = max(sell1, price - buy1) buy2 = min(buy2, price - sell1) sell2 = max(sell2, price - buy2) return sell2 ``` 这个算法的时间复杂度是O(n),其中n是股票价格列表的长度。
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