【leetcode】188. Best Time to Buy and Sell Stock IV

最新推荐文章于 2024-01-20 11:33:56 发布

ghscarecrow

最新推荐文章于 2024-01-20 11:33:56 发布

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本文介绍了一种使用动态规划算法来解决股票交易中寻找最大利润的问题。假设数组中的第i个元素是在第i天某支股票的价格，设计算法找出在最多进行k次交易的情况下，可以获得的最大利润。文章通过具体示例解释了算法的实现过程。

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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.

解题思路：利用动态规划思想。

假设dp[i][j]为在1—j+1天内最多交易i次所取得收益最大值,则

dp[i][j] = max{dp[i][j-1],temp + prices[j]}

temp = {temp,dp[i-1][j-1] + prices[j]}

class Solution {
    public int maxProfit(int k, int[] prices) {
        if(prices.length <= 0 || k <= 0) {
            return 0;
        }
        if(k >= prices.length / 2) {
            return quilkSolve(prices);
        }
        
        int[][] dp = new int[k+1][prices.length];
        
        for(int i = 1;i <= k;i++) {
            int tempMax = - prices[0];
            for(int j = 1;j<prices.length;j++) {
                dp[i][j] = Math.max(dp[i][j - 1],prices[j] + tempMax);
                tempMax = Math.max(tempMax,dp[i - 1][j - 1] - prices[j]);
            }
        }
        
        return dp[k][prices.length - 1];
    }
    
    public int quilkSolve(int[] prices) {
        int maxProfit = 0;
        for(int i = 1;i < prices.length;i++) {
            if(prices[i] > prices[i - 1]) {
                maxProfit += prices[i] - prices[i - 1];
            }
        }
        return maxProfit;
    }
}