LeetCode 188. Best Time to Buy and Sell Stock IV

题目:

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 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:

首先我考虑了动态规划，但是我采用的公式为dp[k][num] = argmaxi dp[k-1][i]+ones[i][num]。(dp[i][j]代表在j日前i项交易的最大值，ones[i][j]代表i~j日一次交易的最大值)事实上这么做复杂度是非常高的，果不其然，超时，下面是超时的代码：

class Solution {public:    int maxProfit(int k, vector<int>& prices) {        if (prices.size() <= 1)return 0;        else if (k == 0)return 0;        else if (k + 1 >= prices.size())        {            int sum = 0;            for (int index = 0; index < prices.size() - 1; index++)            {                sum += maxProfit_one(prices, index, index + 1);            }            return sum;        }        vector<vector<int>> dp(k, vector<int>(prices.size(),0));        vector<vector<int>> ones(prices.size(), vector<int>(prices.size(), 0));        for (int k_index = 1; k_index <= prices.size() - k; k_index++)            for (int index = 0; index < prices.size() - k_index; index++)            {                ones[index][index + k_index] = maxProfit_one(prices, index, index + k_index);            }        for (int index = 1; index < prices.size(); index++)        {            dp[0][index] = ones[0][index];        }        for (int num_k = 1; num_k < k; num_k++)            for (int index = 2; index < prices.size(); index++)            {                dp[num_k][index] = max(dp[num_k - 1][index - 1] + ones[index - 1][index],dp[]);                //cout << num_k << "," << index << ":" << max_value<<endl;            }        return dp[k - 1][prices.size()-1];    }    //[left,right]    int maxProfit_one(vector<int>& prices, int left, int right)    {        int now_num = prices[left];        int min_num = now_num;        int max_money = 0;        int money = 0;        for (int index = left; index <= right; index++)        {            if (prices[index] > now_num)            {                money += (prices[index] - now_num);                now_num = prices[index];            }            else if (min_num > prices[index])            {                max_money = max(max_money, money);                money = 0;                min_num = prices[index];                now_num = min_num;            }            else continue;        }        max_money = max(max_money, money);        return max_money;    }};

之后我又测试了各种方法，要不出错，要不超时…最后AC的思路如下：

• 用二维数组dp[i][j]存储第0~j日最多i次操作所得到的最大利润。

• 用二维数组dps[i][j]存储第0~j日最多i次操作（第2*j次操作为卖出prices[i]）所得到的最大利润。

• 用一维数组ones[i]存储 prices[i+1] > prices[i] ? prices[i+1] - prices[i] : 0

• dps[i][j] = max(dps[i][j - 1] + prices[j] - prices[j - 1], dp[i - 1][j - 1] + ones[j - 1])；

• dp[i][j] = max(dps[i][j],dp[num_k][i]);

class Solution {public:    int maxProfit(int k, vector<int>& prices) {        if (prices.size() <= 1)return 0;        else if (k == 0)return 0;        else if (k + 1 >= prices.size())        {            int sum = 0;            for (int index = 0; index < prices.size() - 1; index++)            {                if (prices[index] < prices[index + 1])                    sum += prices[index + 1] - prices[index];            }            return sum;        }        vector<vector<int>> dp(k+1, vector<int>(prices.size(),0));        vector<vector<int>> dps(k+1, vector<int>(prices.size(), 0));        vector<int> ones(prices.size() - 1, 0);        for (int index = 0; index < prices.size() - 1; index++)        {            if (prices[index] < prices[index + 1])                ones[index] = prices[index + 1] - prices[index];        }        for (int index = 1; index <= k; index++)        {            dp[index][1] = ones[0];            dps[index][1] = dp[index][1];        }        for (int num_k = 1; num_k <= k; num_k++)            for (int index = 2; index < prices.size(); index++)            {                dps[num_k][index] = max(dps[num_k][index - 1] + prices[index] - prices[index - 1], dp[num_k - 1][index - 1] + ones[index - 1]);                dp[num_k][index] = max(dps[num_k][index],dp[num_k][index - 1]);            }        return dp[k][prices.size()-1];    }};