Given an integer array prices, its i-th element prices[i] is the price of a given stock on the i-th day.
Design an algorithm to calculate the maximum profit you can get. You can complete up to k transactions.
Note: You cannot participate in multiple transactions at the same time (you must sell the previous stocks before buying again).
Example 1:
Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on the 1st day (stock price = 2), and sell on the 2nd day (stock price = 4). Profit from this exchange = 4-2 = 2.
Example 2:
Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (stock price = 2), and on day 3 (stock price = 6 ) When you sell, the exchange can make a profit = 6-2 = 4.
Then, buy on the 5th day (stock price = 0) and sell on the 6th day (stock price = 3). The exchange can make a profit = 3-0 = 3.
prompt:
0 <= k <= 109
0 <= prices.length <= 1000
0 <= prices[i] <= 1000
Source: LeetCode
Link: https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv. The
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Idea: I should have written 1, 2, 3 before. The ideas for this question are all in the notes.
class Solution {
public int maxProfit(int k, int[] prices) {
if (prices.length == 0) {
return 0;
}
int n = prices.length;
//最多执行n/2笔交易，买花费一天，卖花费一天
k = Math.min(k, n / 2);
int[][] buy = new int[n][k + 1];//手上有股票
int[][] sell = new int[n][k + 1];//手上没股票
buy[0][0] = -prices[0];
sell[0][0] = 0;
for (int i = 1; i <= k; ++i) {
//这里取MIN_VALUE / 2，3，4，5，6都行，但是不能是MIN_VALUE
//因为MIN_VALUE再减就变成Max_VALUE了，会导致错误
buy[0][i] = sell[0][i] = Integer.MIN_VALUE / 4;
}
//System.out.println(Integer.MIN_VALUE / 2);
//System.out.println(Integer.MIN_VALUE -5 );
for (int i = 1; i < n; ++i) {
//后面天数中从未有过交易的，从前一天就有股票和前一天没有股票买今天的选max
buy[i][0] = Math.max(buy[i - 1][0], sell[i - 1][0] - prices[i]);
for (int j = 1; j <= k; ++j) {
buy[i][j] = Math.max(buy[i - 1][j], sell[i - 1][j] - prices[i]);
sell[i][j] = Math.max(sell[i - 1][j], buy[i - 1][j - 1] + prices[i]);
}
}
//不一定是完成k笔交易的利润最大，比如递减序列。所以从sell【n-1】中取max
return Arrays.stream(sell[n - 1]).max().getAsInt();
}
}