leetcode188-- Best Time to Buy and Sell Stock IV

### 题目

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.





class Solution(object):
def maxProfit(self, k, prices):
"""
:type k: int
:type prices: List[int]
:rtype: int
"""

# 这个规定最多交易两次，即两次买， 两次卖

# local[i][j] 代表在最多进行j次交易的情况下，并且在第i天当天卖掉之后的最大利润
# global[i][j] 代表在最多进行j次交易的情况下，第i天结束时的最大利润

# 则有 l[i][j] = max(g[i-1][j-1] + max(diff, 0), l[i-1][j]+diff)
# g[i][j] = max(l[i][j], g[i-1][j])

if (len(prices)) ==0:
return 0

n = len(prices)

l = [[0 for _ in range(k+1)] for _ in range(n)]   # n行k+1列
g = [[0 for _ in range(k+1)] for _ in range(n)]   # n行k+1列

for i in range(1, len(prices)):
diff = prices[i]- prices[i-1]   # 昨天买今天卖的差值
for j in range(1, k+1):
l[i][j] = max(g[i-1][j-1]+max(diff, 0), l[i-1][j]+diff)
# 因为有两种情况，首先今天必段卖，所以后面的，另外是如是要diff>0的话，如果diff<0的话，就不记了，因为可能还是之前的最大
# 注意虽然今天要卖，但是l代表的是前面这几天中最大的利润

g[i][j] = max(l[i][j], g[i-1][j])  # 更新之前的最大和当前的最大

return g[n-1][k]  