Best Time to Buy and Sell Stock

Best Time to Buy and Sell Stock II

1. 题目描述

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).

2. 思路分析

这题比较简单，因为可以无限次进行买和卖的操作，利用贪心算法的思想：当下一天的价格高于当天的价格，则今天买入股票，明天将股票卖出，这样就能保证得到最大收益。

3. 代码实现

class Solution {public:    int maxProfit(vector<int>& prices) {        int profit = 0;        if (prices.size() <= 1) {            return 0;        }        for (int i = 0; i < prices.size()-1; i++) {            if (prices[i+1] > prices[i]) {                profit += prices[i+1] - prices[i];            }        }        return profit;    }};

Best Time to Buy and Sell Stock IV

1. 题目描述

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).

2. 思路分析

这道题相比较Best Time to Buy and Sell Stock II难度剧增。有了最多k次交易次数的限制条件，因此需要用到动态规划的思想，但是普通的动态规划是行不通的。参考其他博客，定义两个数组local[i][j]：第i 天之前最多进行j 次交易且最后1次交易在最后1天卖出的最大收益，global[i][j]：第i 天之前最多进行j 次交易的最大收益。
对于global[i][j]（最后1天要么有交易，要么没有交易），递推式是：global[i][j] = max(local[i][j], global[i - 1][j])；
对于local[i][j]（倒数第2天要么没有交易，要么有交易），递推式是：local[i][j] = max(global[i - 1][j - 1] + max(diff, 0), local[i - 1][j] + diff)，其中diff = prices[i] - prices[i - 1]。

3. 代码实现

class Solution {public:    int maxProfit(int k, vector<int>& prices) {        int profit = 0;        int n = prices.size();        if (k == 0 || n == 0)            return 0;        else if (k >= n) {            return subMaxProfit(prices);        }        int local[n+2] = {0};        int global[n+2] = {0};        int diffPrice = 0;        for (int i = 0; i < n-1; i++) {            diffPrice = prices[i+1] - prices[i];            for (int j = k; j >= 1; j--) {                local[j] = max(global[j-1] + max(diffPrice, 0), local[j] + diffPrice);              //最后一天没有任何交易 OR 最后一天完成第j次交易              global[j] = max(global[j], local[j]);          }        }        return global[k];    }    int subMaxProfit(vector<int>& prices) {        int profit = 0;        if (prices.size() == 1)             return 0;        for (int i = 0; i < prices.size()-1; i++) {            if (prices[i+1] > prices[i]) {                profit += prices[i+1] - prices[i];            }        }        return profit;    }};