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Array - 188. Best Time to Buy and Sell Stock IV

Array - 188. Best Time to Buy and Sell Stock IV

188. Best Time to Buy and Sell Stock IV

Say you have an array for which the _i_th 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.

思路:

如果会做123题,这一题就一定能想到这个最优解,也是动态规划,题目是允许交易k次,而123题是只允许两次。

代码:

java:

class Solution {

    public int maxProfit(int k, int[] prices) {
        int len = prices.length;
        if (len == 0) return 0;
        if (k >= len/2) return helper(prices);
        
        int[] dp = new int[k+1];
        int[] min = new int[k+1];
        for (int i = 0; i <min.length; i++) {
            min[i] = prices[0];
        }
        
        for (int i = 0; i < prices.length; i++) {
            for (int j = 1; j <= k; j++) {
                dp[j] = Math.max(dp[j], prices[i] - min[j]);
                min[j] = Math.min(min[j], prices[i] - dp[j-1]);
            }
        }
        return dp[k];
    }
    
    private int helper(int[] p) {
        int len = p.length;
        int res = 0;
        
        for(int i = 1; i < len; i++) {
            if (p[i] > p[i-1]) res += (p[i] - p[i-1]);
        }
        
        return res;
    }
}

标题:Array - 188. Best Time to Buy and Sell Stock IV
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/06/16/1560678912044.html

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