Leetcode Best time to buy and sell stock 问题

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本文详细介绍了三种不同的股票交易策略：一次交易、无限次交易和最多两次交易的情况。通过代码实现，解释了如何在有限或无限交易次数下最大化利润。

Best time to buy and sell stock I

Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

[code]

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

Best time to buy and sell stock II

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

[code]

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices==null || prices.length<=1)return 0;
        int profit=0;
        for(int i=1;i<prices.length;i++)
        {
            profit+=Math.max(0,prices[i]-prices[i-1]);
        }
        return Math.max(0,profit);
    }
}

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

[code]

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices==null || prices.length<2)return 0;
        int max[]=new int[prices.length], min[]=new int [prices.length];
        for(int i=prices.length-1;i>=0;i--)
        {
            if(i==prices.length-1)max[i]=prices[i];
            else max[i]=Math.max(max[i+1],prices[i]);
        }
        for(int i=0;i<prices.length;i++)
        {
            if(i==0)min[i]=prices[i];
            else min[i]=Math.min(min[i-1], prices[i]);
        }
        int d1[]=new int[prices.length], d2[]=new int[prices.length];
        for(int i=0;i<prices.length;i++)
        {
            if(i==0)d1[0]=0;
            else
            {
                d1[i]=Math.max(d1[i-1], prices[i]-min[i]);
            }
        }
        for(int i=prices.length-1;i>=0;i--)
        {
            if(i==prices.length-1)d2[i]=0;
            else
            {
                d2[i]=Math.max(d2[i+1], max[i]-prices[i]);
            }
        }
        int profit=0;
        for(int i=0;i<prices.length;i++)profit=Math.max(profit, d1[i]+d2[i]);
        return profit;
    }
}

[Thought]
划分成2段分别进行dp