[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).

解题思路

采用动态规划来解决问题。

我们需要维护如下两个量:

global[i][j]:当前到达第i天最多可以进行j次交易,所得到的最大利润。

local[i][j]:当前到达第i天最多可以进行j次交易,而且最后一次交易在当天卖出,所得到的最大利润。

状态转移方程:

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

上述方程比较两个量的大小:①当前局部最大值;②过往全局最大值。

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

上述方程比较两个量的大小:

①全局到i-1天进行j-1次交易,然后加上今天的交易(如果今天的交易赚钱的话)。

②取局部第i-1天进行j次交易,然后加上今天的差值(local[i-1][j]是第i-1天卖出的交易,它加上diff后变成第i天卖出,并不会增加交易次数。无论diff是正还是负都要加上,否则就不满足local[i][j]必须在最后一天卖出的条件了)

另外需要注意的一个问题是,当k远大于数组的大小时,上述算法将变得低效。因此将其改用不限交易次数的方式解决。

实现代码

/*****************************************************************
    *  @Author   : 楚兴
    *  @Date     : 2015/2/22 23:42
    *  @Status   : Accepted
    *  @Runtime  : 15 ms
******************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    int maxProfit(int k, vector<int> &prices) {
        int len = prices.size();
        if (len == 0)
        {
            return 0;
        }
        if (k >= len)
        {
            return helper(prices);
        }

        vector<int> max_local(k + 1, 0);
        vector<int> max_global(k + 1, 0);
        int diff;
        for (int i = 0; i < len - 1; i++)
        {
            diff = prices[i + 1] - prices[i];
            for (int j = k; j >= 1; j--)
            {
                max_local[j] = max(max_global[j - 1] + max(diff, 0), max_local[j] + diff);
                max_global[j] = max(max_local[j], max_global[j]);
            }
        }

        return max_global[k];
    }

    int helper(vector<int> &prices)
    {
        int profit = 0;
        for (int i = 0; i < prices.size() - 1; i++)
        {
            profit = max(profit, profit + prices[i + 1] - prices[i]);
        }

        return profit;
    }
};
时间: 2024-11-29 00:21:56

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