LeetCode 188 Best Time to Buy and Sell Stock IV (动态规划 推荐)

Say you have an array for which the i^th element is the price of a given stock on dayi.

Design an algorithm to find the maximum profit. You may complete at mostk transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

题目链接：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/

题目分析：股票系列最后一题，最多买k次，对于两次的时候有这样一种很好理解的解法：

for (int price : prices) {    buy1 = Math.max(buy1, -price);    sell1 = Math.max(sell1, buy1 + price);    buy2 = Math.max(buy2, sell1 - price);    sell2 = Math.max(sell2, buy2 + price);}

然后推广的k次，实际上就是加一层循环，有一个优化，如果k >= n / 2，相当于可以买无限次，直接用无限次的做法On扫一下即可

public class Solution {    public int maxProfit(int k, int[] prices) {        int n = prices.length;        if (n < 2) {            return 0;        }        if (k >= n / 2) {            int ans = 0;            for (int i = 0; i < n - 1; i ++) {                if (prices[i + 1] > prices[i]) {                    ans += prices[i + 1] - prices[i];                }            }            return ans;        }        else {            int[] sell = new int[n + 1];            int[] buy = new int[n + 1];            for (int i = 1; i <= n; i ++) {                buy[i] = -0x3fffffff;            }            for (int i = 0; i < n; i ++) {                for (int j = 1; j <= k; j ++) {                    buy[j] = Math.max(buy[j], sell[j - 1] - prices[i]);                    sell[j] = Math.max(sell[j], buy[j] + prices[i]);                }            }            return sell[k];         }    }}