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# 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. ~~~ ### 題解1 賣股票系列中最難的一道，較易實(shí)現(xiàn)的方法為使用動態(tài)規(guī)劃，動規(guī)的實(shí)現(xiàn)又分為大約3大類方法，這里先介紹一種最為樸素的方法，過不了大量數(shù)據(jù)，會 [TLE](# "Time Limit Exceeded 的簡稱。你的程序在 OJ 上的運(yùn)行時(shí)間太長了，超過了對應(yīng)題目的時(shí)間限制。"). 最多允許 k 次交易，由于一次增加收益的交易至少需要兩天，故當(dāng) k >= n/2時(shí)，此題退化為賣股票的第二道題，即允許任意多次交易。當(dāng) k < n/2 時(shí)，使用動規(guī)來求解，動規(guī)的幾個(gè)要素如下： f[i][j] 代表第 i 天為止交易 k 次獲得的最大收益，那么將問題分解為前 x 天交易 k-1 次，第 x+1 天至第 i 天交易一次兩個(gè)子問題，于是動態(tài)方程如下： ~~~ f[i][j] = max(f[x][j - 1] + profit(x + 1, i)) ~~~ 簡便起見，初始化二維矩陣為0，下標(biāo)盡可能從1開始，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; } }; ~~~ ### 源碼分析 注意 Python 中的多維數(shù)組初始化方式，不可簡單使用`[[0] * k] * n]`, 具體原因是因?yàn)?Python 中的對象引用方式?？梢詢?yōu)化的地方是 profit 方法及最內(nèi)存循環(huán)。 ### 復(fù)雜度分析 三重循環(huán)，時(shí)間復(fù)雜度近似為 O(n2?k)O(n^2 \cdot k)O(n2?k), 使用了 f 二維數(shù)組，空間復(fù)雜度為 O(n?k)O(n \cdot k)O(n?k). ### Reference - [[LeetCode] Best Time to Buy and Sell Stock I II III IV | 梁佳賓的網(wǎng)絡(luò)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 參考程序 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 // 陳輝的技術(shù)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 | 書影博客](http://bookshadow.com/weblog/2015/02/18/leetcode-best-time-to-buy-and-sell-stock-iv/)