188. Best Time to Buy and Sell Stock IV

You are given an integer array prices where prices[i] is the price of a given stock on the ith day.

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

Notice that you may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: k = 2, prices = [3,2,6,5,0,3]
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.

Constraints:

  • 0 <= k <= 109
  • 0 <= prices.length <= 104
  • 0 <= prices[i] <= 1000

dp (optimal), time = O(nk), space = O(k)

class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if(n < 2) {
            return 0;
        }
        if(k >= n / 2) {
            int maxProfit = 0;
            for(int i = 1; i < n; i++) {
                maxProfit += Math.max(0, prices[i] - prices[i - 1]);
            }
            return maxProfit;
        }
        
        int[] local = new int[k + 1];   // local[i]: the max profit on day i, stock must be sold on day i
        int[] global = new int[k + 1]; // global[i]: the max profit on day i

        for(int i = 1; i < n; i++) {
            int diff = prices[i] - prices[i - 1];
            for(int j = k; j >= 1; j--) {
                local[j] = Math.max(global[j - 1] + Math.max(0, diff), local[j] + diff);
                global[j] = Math.max(global[j], local[j]);
            }
        }
        return global[k];
    }
}

dp, time = O(nk), space = O(nk)

class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if(n < 2) {
            return 0;
        }
        if(k >= n / 2) {
            int maxProfit = 0;
            for(int i = 1; i < n; i++) {
                maxProfit += Math.max(0, prices[i] - prices[i - 1]);
            }
            return maxProfit;
        }
        
        // mustSell[i][j]: the max profit in first i days with at most j transactions, stock must be sold on day i
        // globalBest[i][j]: the max profit in first i days with at most j transactions
        int[][] mustSell = new int[n][k + 1];
        int[][] globalBest = new int[n][k + 1];
        
        for(int i = 1; i < n; i++) {
            int diff = prices[i] - prices[i - 1];
            mustSell[i][0] = 0;
            for(int j = 1; j <= k; j++) {
                mustSell[i][j] = Math.max(globalBest[i - 1][j - 1] + diff, mustSell[i - 1][j] + diff);
                globalBest[i][j] = Math.max(globalBest[i - 1][j], mustSell[i][j]);
            }
        }
        return globalBest[n - 1][k];
    }
}
原文地址:https://www.cnblogs.com/fatttcat/p/13835998.html