原题传送门

1. 题目描述

2. Solution 1

1、思路分析
1> dp[i, j] represents the max profit up until prices[j] using at most i transactions.
2> dp[0, j] = 0; 0 transactions makes 0 profit
dp[i, 0] = 0; if there is only one price data point you can't make any transaction.
3> dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] }
= max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj]))

2、代码实现

package Q0199.Q0188BestTimetoBuyandSellStockIV;

/*
  dp[i, j] represents the max profit up until prices[j] using at most i transactions.
  dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] }
           = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj]))
  dp[0, j] = 0; 0 transactions makes 0 profit
  dp[i, 0] = 0; if there is only one price data point you can't make any transaction.
 */
public class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if (n <= 1) return 0;

        //if k >= n/2, then you can make maximum number of transactions.
        if (k >= n / 2) {
            int maxPro = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] > prices[i - 1])
                    maxPro += prices[i] - prices[i - 1];
            }
            return maxPro;
        }

        int[][] dp = new int[k + 1][n];
        for (int i = 1; i <= k; i++) {
            int localMax = dp[i - 1][0] - prices[0];
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.max(dp[i][j - 1], prices[j] + localMax);
                localMax = Math.max(localMax, dp[i - 1][j] - prices[j]);
            }
        }
        return dp[k][n - 1];
    }
}

3、复杂度分析
时间复杂度: O(kn)
空间复杂度: O(kn)

标签：Sell,Buy,transactions,int,max,0188,jj,prices,dp
来源： https://www.cnblogs.com/junstat/p/16325561.html