[LeetCOde][Java] Best Time to Buy and Sell Stock III

標籤：leetcode   java   best time to buy and   

題目：

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).

演算法分析：

利用通用的演算法《Best Time to Buy and Sell Stock IV》 將k變為2 就ok 

AC代碼：

<span style="font-family:Microsoft YaHei;font-size:12px;">public class Solution{    public int maxProfit(int[] prices)     {        return maxProfit(2,prices);    }    public int maxProfit(int k, int[] prices)     {       if(prices==null || prices.length==0)            return 0;        if(k>prices.length)//k次數大於天數時，轉化為問題《Best Time to Buy and Sell Stock II》--無限次交易的情景         {            if(prices==null)                return 0;            int res=0;            for(int i=0;i<prices.length-1;i++)            {                int degit=prices[i+1]-prices[i];                if(degit>0)                    res+=degit;            }            return res;        }        /*      定義維護量：      global[i][j]:在到達第i天時最多可進行j次交易的最大利潤，此為全域最優      local[i][j]:在到達第i天時最多可進行j次交易並且最後一次交易在最後一天賣出的最大利潤，此為局部最優      定義遞推式：      global[i][j]=max(global[i-1][j],local[i][j]);即第i天沒有交易，和第i天有交易      local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff)  diff=price[i]-price[i-1];       */        int[][] global=new int[prices.length][k+1];        int[][] local=new int[prices.length][k+1];        for(int i=0;i<prices.length-1;i++)        {            int diff=prices[i+1]-prices[i];            for(int j=0;j<=k-1;j++)            {                local[i+1][j+1]=Math.max(global[i][j]+Math.max(diff,0),local[i][j+1]+diff);                global[i+1][j+1]=Math.max(global[i][j+1],local[i+1][j+1]);            }        }        return global[prices.length-1][k];    }}</span>

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[LeetCOde][Java] Best Time to Buy and Sell Stock III