Best Time to Buy and Sell Stock IV
最后更新于：2022-04-02 01:13:20
# 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 &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 > f = vector >(n + 1, vector(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 &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/)
';