Best Time to Buy and Sell Stock IV · 数据结构与算法/leetcode/lintcode题解

# Best Time to Buy and Sell Stock IV ### Source - leetcode: [Best Time to Buy and Sell Stock IV | LeetCode OJ](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/) - lintcode: [(393) Best Time to Buy and Sell Stock IV](http://www.lintcode.com/en/problem/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. Example Given prices = [4,4,6,1,1,4,2,5], and k = 2, return 6. Note You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again). Challenge O(nk) time. ~~~ ### 题解1 卖股票系列中最难的一道，较易实现的方法为使用动态规划，动规的实现又分为大约3大类方法，这里先介绍一种最为朴素的方法，过不了大量数据，会 [TLE](# "Time Limit Exceeded 的简称。你的程序在 OJ 上的运行时间太长了，超过了对应题目的时间限制。"). 最多允许 k 次交易，由于一次增加收益的交易至少需要两天，故当 k >= n/2时，此题退化为卖股票的第二道题，即允许任意多次交易。当 k < n/2 时，使用动规来求解，动规的几个要素如下： f[i][j] 代表第 i 天为止交易 k 次获得的最大收益，那么将问题分解为前 x 天交易 k-1 次，第 x+1 天至第 i 天交易一次两个子问题，于是动态方程如下： ~~~ f[i][j] = max(f[x][j - 1] + profit(x + 1, i)) ~~~ 简便起见，初始化二维矩阵为0，下标尽可能从1开始，便于理解。 ### Python ~~~ class Solution: """ @param k: an integer @param prices: a list of integer @return: an integer which is maximum profit """ def maxProfit(self, k, prices): if prices is None or len(prices) <= 1 or k <= 0: return 0 n = len(prices) # k >= prices.length / 2 ==> multiple transactions Stock II if k >= n / 2: profit_max = 0 for i in xrange(1, n): diff = prices[i] - prices[i - 1] if diff > 0: profit_max += diff return profit_max f = [[0 for i in xrange(k + 1)] for j in xrange(n + 1)] for j in xrange(1, k + 1): for i in xrange(1, n + 1): for x in xrange(0, i + 1): f[i][j] = max(f[i][j], f[x][j - 1] + self.profit(prices, x + 1, i)) return f[n][k] # calculate the profit of prices(l, u) def profit(self, prices, l, u): if l >= u: return 0 valley = 2**31 - 1 profit_max = 0 for price in prices[l - 1:u]: profit_max = max(profit_max, price - valley) valley = min(valley, price) return profit_max ~~~ ### C++ ~~~ class Solution { public: /** * @param k: An integer * @param prices: Given an integer array * @return: Maximum profit */ int maxProfit(int k, vector<int> &prices) { if (prices.size() <= 1 || k <= 0) return 0; int n = prices.size(); // k >= prices.length / 2 ==> multiple transactions Stock II if (k >= n / 2) { int profit_max = 0; for (int i = 1; i < n; ++i) { int diff = prices[i] - prices[i - 1]; if (diff > 0) { profit_max += diff; } } return profit_max; } vector<vector<int> > f = vector<vector<int> >(n + 1, vector<int>(k + 1, 0)); for (int j = 1; j <= k; ++j) { for (int i = 1; i <= n; ++i) { for (int x = 0; x <= i; ++x) { f[i][j] = max(f[i][j], f[x][j - 1] + profit(prices, x + 1, i)); } } } return f[n][k]; } private: int profit(vector<int> &prices, int l, int u) { if (l >= u) return 0; int valley = INT_MAX; int profit_max = 0; for (int i = l - 1; i < u; ++i) { profit_max = max(profit_max, prices[i] - valley); valley = min(valley, prices[i]); } return profit_max; } }; ~~~ ### Java ~~~ class Solution { /** * @param k: An integer * @param prices: Given an integer array * @return: Maximum profit */ public int maxProfit(int k, int[] prices) { if (prices == null || prices.length <= 1 || k <= 0) return 0; int n = prices.length; if (k >= n / 2) { int profit_max = 0; for (int i = 1; i < n; i++) { if (prices[i] - prices[i - 1] > 0) { profit_max += prices[i] - prices[i - 1]; } } return profit_max; } int[][] f = new int[n + 1][k + 1]; for (int j = 1; j <= k; j++) { for (int i = 1; i <= n; i++) { for (int x = 0; x <= i; x++) { f[i][j] = Math.max(f[i][j], f[x][j - 1] + profit(prices, x + 1, i)); } } } return f[n][k]; } private int profit(int[] prices, int l, int u) { if (l >= u) return 0; int valley = Integer.MAX_VALUE; int profit_max = 0; for (int i = l - 1; i < u; i++) { profit_max = Math.max(profit_max, prices[i] - valley); valley = Math.min(valley, prices[i]); } return profit_max; } }; ~~~ ### 源码分析注意 Python 中的多维数组初始化方式，不可简单使用`[[0] * k] * n]`, 具体原因是因为 Python 中的对象引用方式。可以优化的地方是 profit 方法及最内存循环。 ### 复杂度分析三重循环，时间复杂度近似为 O(n2⋅k)O(n^2 \cdot k)O(n2⋅k), 使用了 f 二维数组，空间复杂度为 O(n⋅k)O(n \cdot k)O(n⋅k). ### Reference - [[LeetCode] Best Time to Buy and Sell Stock I II III IV | 梁佳宾的网络日志](http://liangjiabin.com/blog/2015/04/leetcode-best-time-to-buy-and-sell-stock.html) - [Best Time to Buy and Sell Stock IV 参考程序 Java/C++/Python](http://www.jiuzhang.com/solutions/best-time-to-buy-and-sell-stock-iv/) - [leetcode-Best Time to Buy and Sell Stock 系列 // 陈辉的技术博客](http://www.devhui.com/2015/02/23/Best-Time-to-Buy-and-Sell-Stock/) - [[LeetCode]Best Time to Buy and Sell Stock IV | 书影博客](http://bookshadow.com/weblog/2015/02/18/leetcode-best-time-to-buy-and-sell-stock-iv/)