算法: 买卖股票的时间 121. Best Time to Buy and Sell Stock

Posted 2021-12-15 架构师易筋

121. Best Time to Buy and Sell Stock

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

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 104

Kadane 的算法

解决这个问题的逻辑与使用Kadane’s Algorithm. 由于到目前为止还没有任何机构提到这一点，我认为每个人都知道这是一件好事。

所有直接的解决方案都应该有效，但是如果面试官通过给出价格差异数组来稍微扭曲问题，例如：for 1, 7, 4, 11，如果他给出0, 6, -3, 7，你最终可能会感到困惑。

这里的逻辑是计算maxCur += prices[i] - prices[i-1]原始数组的差值 ( )，并找到一个利润最大的连续子数组。如果差值低于 0，则将其重置为零。

• maxCur = current maximum value

• maxSoFar = maximum value found so far

class Solution 
    public int maxProfit(int[] prices) 
        int maxCur = 0, maxSofar = 0;
        for (int i = 1; i < prices.length; i++) 
            maxCur += prices[i] - prices[i - 1];
            maxCur = Math.max(0, maxCur);
            maxSofar = Math.max(maxSofar, maxCur);
        
        return maxSofar;
    

参考

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/discuss/39038/Kadane’s-Algorithm-Since-no-one-has-mentioned-about-this-so-far-%3A)-(In-case-if-interviewer-twists-the-input)