Best Time to Buy and Sell Stock with Cooldown

描述

Almost the ame as Best Time to Buy and Sell Stock II but with one restriction: after you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day).

Example:

prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]

分析

这题比Best Time to Buy and Sell Stock II多了一个cooldown的条件,就变得麻烦多了。这题是一个多阶段优化问题,首先范围缩小到广搜,贪心或者动规。因为每步之间互相牵连,贪心显然不行。广搜固然可以,不过是O(2^n)复杂度,所以我们先考虑用动规。

对于每一天,有三种动作,buy, sell, cooldown, sell 和 cooldown 可以合并成一种状态,因为手里最终没有股票。最终需要的结果是 sell,即手里股票卖了获得最大利润。我们可以用两个数组来记录当前持股和未持股的状态,令sell[i] 表示第i天未持股时,获得的最大利润,buy[i]表示第i天持有股票时,获得的最大利润。

对于sell[i],最大利润有两种可能,一是今天没动作跟昨天未持股状态一样,二是今天卖了股票,所以状态转移方程如下:

sell[i] = max{sell[i - 1], buy[i-1] + prices[i]}

对于buy[i],最大利润有两种可能,一是今天没动作跟昨天持股状态一样,二是前天卖了股票,今天买了股票,因为 cooldown 只能隔天交易,所以今天买股票要追溯到前天的状态。状态转移方程如下:

buy[i] = max{buy[i-1], sell[i-2] - prices[i]}

最终我们要求的结果是sell[n - 1],表示最后一天结束时,手里没有股票时的最大利润。

这个算法的空间复杂度是O(n),不过由于sell[i]仅仅依赖前一项,buy[i]仅仅依赖前两项,所以可以优化到O(1),具体见第二种代码实现。

代码1 O(n)空间

// Best Time to Buy and Sell Stock with Cooldown
// Time Complexity: O(n), Space Complexity: O(n)
public class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) return 0;

        int[] sell = new int[prices.length];
        int[] buy = new int[prices.length];
        sell[0] = 0;
        buy[0] = -prices[0];

        for (int i = 1; i < prices.length; ++i) {
            sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]);
            buy[i] = Math.max(buy[i - 1], (i > 1 ? sell[i - 2] : 0) - prices[i]);
        }
        return sell[prices.length - 1];
    }
}

代码2 O(1)空间

// Best Time to Buy and Sell Stock with Cooldown
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) return 0;

        int curSell = 0;   // sell[i]
        int prevSell = 0;  // sell[i-2]
        int buy = -prices[0]; // buy[i]

        for (int i = 1; i < prices.length; ++i) {
            final int tmp = curSell;
            curSell = Math.max(curSell, buy + prices[i]);
            buy = Math.max(buy, (i > 1 ? prevSell : 0) - prices[i]);
            prevSell = tmp;
        }
        return curSell;
    }
}

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