LeetCode Best Time to Buy and Sell Stock IV-优快云博客

本文针对LeetCode上的一道经典动态规划题目“Best Time to Buy and Sell Stock IV”提供了详细的解答思路与Java代码实现。文章介绍了如何根据交易次数限制设计算法以求得最大利润，并考虑了特殊情况下的简化处理。

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原题链接在这里：https://leetcode.com/problems/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).

题解：

本题是Best Time to Buy and Sell Stock III的general情况。

但这里要注意若是k很大已经超过了prices.length的时候, 若是按照DP的方法做会浪费时间以及空间. 此时可以直接参照Best Time to Buy and Sell Stock II做法.

Time Complexity: O(prices.length * k), k是最多交易次数.

Space: O(k).

AC Java:

 1 public class Solution {
 2     public int maxProfit(int k, int[] prices) {
 3         if(prices == null || prices.length == 0 || k == 0){
 4             return 0;
 5         }
 6         if(k>=prices.length/2){
 7             int res = 0;
 8             for(int i = 1; i < prices.length; i++){
 9                 res += Math.max(0, prices[i]-prices[i-1]);
10             }
11             return res;
12         }
13         
14         int[] local = new int[k+1];
15         int[] global = new int[k+1];
16         for(int i = 1; i<prices.length; i++){
17             int diff = prices[i] - prices[i-1];
18             for(int j = k; j>=1; j--){
19                 local[j] = Math.max(global[j-1] + Math.max(diff,0), local[j] + diff);
20                 global[j] = Math.max(global[j], local[j]);
21             }
22         }
23         return global[k];
24     }
25 }