# 188. Best Time to Buy and Sell Stock IV 买卖股票的最佳时机 IV

@TOC

## # 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).

Example 1:

``````Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
``````

Example 2:

``````Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
``````

## # 解题方法

local[i][j] = max(global[i - 1][j - 1] + max(diff, 0), local[i - 1][j] + diff)

global[i][j] = max(local[i][j], global[i - 1][j])，

``````class Solution(object):
def maxProfit(self, k, prices):
"""
:type k: int
:type prices: List[int]
:rtype: int
"""
if k <= 0 or not prices: return 0
N = len(prices)
if k >= N:
_sum = 0
for i in xrange(1, N):
if prices[i] > prices[i - 1]:
_sum += prices[i] - prices[i - 1]
return _sum
g = [0] * (k + 1)
l = [0] * (k + 1)
for i in xrange(N - 1):
diff = prices[i + 1] - prices[i]
for j in xrange(k, 0, -1):
l[j] = max(g[j - 1] + max(diff, 0), l[j] + diff)
g[j] = max(l[j], g[j])
return g[-1]
``````

## # 日期

2018 年 12 月 1 日 —— 2018年余额不足了