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Array - 123. Best Time to Buy and Sell Stock III

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ppxai
发布2020-09-23 17:07:21
发布2020-09-23 17:07:21
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文章被收录于专栏:皮皮星球皮皮星球

123. Best Time to Buy and Sell Stock III

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 two transactions.

**Note: **You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [3,3,5,0,0,3,1,4] Output: 6 Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.   Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.   Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are   engaging multiple transactions at the same time. You must sell before buying again.

思路:

这一题也是买卖股票的问题,条件是允许买卖两次,最直接的办法是在121的基础上,分为两次最大盈利做。这里只说一下动态规划来解。

代码:

java:

代码语言:javascript
复制
class Solution {

    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) return 0;
        
        int[] dp = new int[3];
        int[] min = new int[]{prices[0], prices[0], prices[0]};
        
        for (int i = 0; i < prices.length; i++) {
            for (int k = 1; k <= 2; k++) {
                dp[k] = Math.max(dp[k], prices[i] - min[k]);
                min[k] = Math.min(min[k], prices[i] - dp[k-1]);
            }
        }
        return dp[2];
    }
}
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原始发表:2019年06月16日,如有侵权请联系 cloudcommunity@tencent.com 删除

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