友情支持
如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜
|
|
有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。
公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。 如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。 |
123. Best Time to Buy and Sell Stock III
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 two transactions.
Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
engaging multiple transactions at the same time. You must sell before buying again.
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
针对 一个方法团灭 6 道股票问题 - 最佳买卖股票时机含冷冻期 - 力扣（LeetCode） 这个解题框架，进行小试牛刀。
思考题：将表格改成变量时，还需要再思考思考。
参考资料
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 two transactions.
*Note: *You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Example 1:
Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:
Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
/**
* Runtime: 2 ms, faster than 47.09% of Java online submissions for Best Time to Buy and Sell Stock III.
* Memory Usage: 42.5 MB, less than 7.32% of Java online submissions for Best Time to Buy and Sell Stock III.
*
* Copy from: https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/solution/yi-ge-fang-fa-tuan-mie-6-dao-gu-piao-wen-ti-by-lab/[一个方法团灭 6 道股票问题 - 最佳买卖股票时机含冷冻期 - 力扣（LeetCode）]
*/
public int maxProfit(int[] prices) {
int dp10 = 0, dp11 = Integer.MIN_VALUE;
int dp20 = 0, dp21 = Integer.MIN_VALUE;
for (int i = 0; i < prices.length; i++) {
dp20 = Math.max(dp20, dp21 + prices[i]);
dp21 = Math.max(dp21, dp10 - prices[i]);
dp10 = Math.max(dp10, dp11 + prices[i]);
dp11 = Math.max(dp11, -prices[i]);
}
return dp20;
}
/**
* Runtime: 5 ms, faster than 22.10% of Java online submissions for Best Time to Buy and Sell Stock III.
* Memory Usage: 42.4 MB, less than 7.32% of Java online submissions for Best Time to Buy and Sell Stock III.
*/
public int maxProfitDpTable(int[] prices) {
if (Objects.isNull(prices) || prices.length == 0) {
return 0;
}
int maxK = 2;
int[][][] dp = new int[prices.length][maxK + 1][2];
for (int i = 0; i < prices.length; i++) {
for (int k = maxK; k >= 1; k--) {
if (i == 0) {
dp[i][k][0] = 0;
dp[i][k][1] = -prices[i];
continue;
}
dp[i][k][0] = Math.max(dp[i - 1][k][0], dp[i - 1][k][1] + prices[i]);
dp[i][k][1] = Math.max(dp[i - 1][k][1], dp[i - 1][k - 1][0] - prices[i]);
}
}
return dp[prices.length - 1][maxK][0];
}