【动态规划】Best Time to Buy and Sell Stock IV

题目：leetcode

Best Time to Buy and Sell Stock IV

 

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 k transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

分析：

我们定义local[i][j]为在到达第i天时最多可进行j次交易并且最后一次交易在最后一天卖出的最大利润，此为局部最优。然后我们定义global[i][j]为在到达第i天时最多可进行j次交易的最大利润，此为全局最优。它们的递推式为：

local[i][j] = max(global[i - 1][j - 1] + max(diff, 0), local[i - 1][j] + diff)

global[i][j] = max(local[i][j], global[i - 1][j])

其中局部最优值是比较前一天并少交易一次的全局最优加上大于0的差值，和前一天的局部最优加上差值后相比，两者之中取较大值，而全局最优比较局部最优和前一天的全局最优。

若k值大于等于天数减一，则可以视为交易次数没有限制，可单独讨论。

class Solution {public:    int maxProfit(int k, vector<int> &prices) {        int size=prices.size();        if(size<=1 && k<=0)        return 0;        if(k>=size-1)        return buxianzhi(prices);                vector<vector<int>> local(size,vector<int>(k+1,0)),global(size,vector<int>(k+1,0));        for(int i=1;i<size;i++)        {            int diff=prices[i]-prices[i-1];            for(int j=1;j<=k;j++)            {                if(j>i)                {                    local[i][j]=local[i][i];                    global[i][j]=global[i][i];                    continue;                }               local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff);               global[i][j]=max(global[i-1][j],local[i][j]);            }        }                return global[size-1][k];    }        int buxianzhi(vector<int> &prices)    {        int res=0,size=prices.size();        for(int i=1;i<size;i++)        {            if(prices[i]>prices[i-1])            res+=prices[i]-prices[i-1];        }        return res;    }};