188. Best Time to Buy and Sell Stock IV

男娘i 2022-02-12 11:27 66阅读 0赞

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.

本题是一个经典的动态规划题目。详情见注解。

public int maxProfit(int k, int[] prices) {
            int n = prices.length;
            if(n <= 1) {//无法完成任何交易
                return 0;
            }
            if(k >= n / 2) {//可以完成最大次数的交易
                int ans = 0;
                for(int i = 1; i < n; ++i) {
                    if(prices[i] > prices[i-1])
                        ans += prices[i] - prices[i-1];
                }
                return ans;
            }
            int[][] dp = new int[k+1][n];
            /**
             * dp[i, j] 表示股市直到第j天交易i次可以获得的最大利润
             * dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) （0<=jj<=j-1）
             *          = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj]))
             * dp[0, j] = 0; 0次交易利润为0
             * dp[i, 0] = 0; 在第一天不能完成一次交易，利润为0
             */
            for(int i = 1; i <= k; ++i) {
                //tempMax表示prices[jj] + dp[i-1][jj])(0<=jj<=j-1)的最大值
                int tempMax = dp[i-1][0] - prices[0];
                for(int j = 1; j < n; ++j) {
                    //dp[i][j] = max(dp[i-1][j], prices[j] - prices[jj] + dp[i-1][jj])(0<=jj<=j-1)
                    dp[i][j] = Math.max(dp[i][j-1], prices[j] + tempMax);
                    tempMax = Math.max(tempMax, dp[i-1][j] - prices[j]);
                }
            }
            return dp[k][n-1];
        }