188、Best Time to Buy and Sell Stock IV

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.

思路

动态规划的方法来做,参考资料1的方法比较晦涩难懂,搜了一下,好像大部分都是采用这种方法来做的,放弃了。

找到了一个好懂一点的方法,学习一下:

类似最大连续和问题,我们需要变量来记录当前的一些状态信息,因为是k次交易,所以我们需要大小为k的数组记录每一次交易的信息,

又因为一次交易包含买和卖两次操作,所以我们需要两个数组。分别命名为hold【】和sell【】。

  • hold[i]表示第i次持有后的盈亏,也就是第i次买入后的状态
  • sell[i]表示第i次卖出后的盈亏状态

比如上个例子1,3,7,2,1,5.如果我们在第一天买入,那么hold[1]应该是-1,这里可能会问,为什么会是负数?那是因为我们买入了股票,钱已经花出去了,但我们还没有卖出,所以当前你手上是没有现金的,只有股票。

算法核心就是两句:

cur = prices[i];
sell[j] = std::max(sell[j], hold[j] + cur);
hold[j] = std::max(hold[j], sell[j-1] - cur);

在第i天第j次 卖出 的盈亏最大值为 已卖出第j次持有并在第i天卖出 之间的最大值。

在第i天第j次 持有 的盈亏最大值为 继续持有第j-1次卖出并在第i天又买入 之间的最大值。

c++11

如果k大于等于长度的一半,则说明可以任意次交易得到最大值。否则采用dp算法。

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if (k >= prices.size()/2) {
            int res = 0;
            for (int i = 1; i < prices.size(); i++)
                if (prices[i] > prices[i-1]) 
                    res += std::max(prices[i] - prices[i-1], 0);
            return res;
        }

        int hold[k+1];
        int sell[k+1];
        for (int i=0; i<=k; ++i){
            hold[i] = INT_MIN;
            sell[i] = 0;
        }

        for (int i = 0; i < prices.size(); i++) {
            for (int j = 1; j <= k; j++) {   
                sell[j] = std::max(sell[j], hold[j] + prices[i]);
                hold[j] = std::max(hold[j], sell[j-1] - prices[i]);   
            }
        }
        return sell[k];
    }
};

AC结果:

Runtime: 8 ms, faster than 100.00% of C++ online submissions for Best Time to Buy and Sell Stock IV.
Memory Usage: 9 MB, less than 61.68% of C++ online submissions for Best Time to Buy and Sell Stock IV.

results matching ""

    No results matching ""