121. Best Time to Buy and Sell Stock-优快云博客

本文链接：https://blog.youkuaiyun.com/daimingyang123/article/details/78610267

本文介绍了一种寻找股票买卖最佳时机以获得最大利润的算法。通过两种方法实现：一种是记录遍历过程中的最低买入价格并计算后续可能出现的最大利润；另一种则采用Kadane算法，适用于处理价格差值数组。

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Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]

Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]

Output: 0

In this case, no transaction is done, i.e. max profit = 0.

之前记得谁说过数据结构与算法题目最难不过二叉树但我总觉得动态规划挺难的

首先想到的这种解法

    public int maxProfit(int[] prices) {
        if (prices.length == 0) return 0;
        
        int min = prices[0];
        int profit = 0;
        for (int i = 1; i < prices.length; i++) {
            if (prices[i] < min) {
                min = prices[i];
            } else {
                if (prices[i] - min > profit) profit = prices[i] - min;
            }
        }
        return profit;
    }

把输入画个图就很好理解了

图片来源https://leetcode.com/problems/best-time-to-buy-and-sell-stock/solution/

我们需要做的就是找到最低的山谷和他后面最高的山顶算出收益求最大受益

维护一个到当前元素为止的最小值min 如果遇到比他小的就算一下收益

第二种解法更加抽象一些

The logic to solve this problem is same as "max subarray problem" using Kadane's Algorithm. Since no body has mentioned this so far, I thought it's a good thing for everybody to know.

All the straight forward solution should work, but if the interviewer twists the question slightly by giving the difference array of prices, Ex: for {1, 7, 4, 11}, if he gives {0, 6, -3, 7}, you might end up being confused.

Here, the logic is to calculate the difference (maxCur += prices[i] - prices[i-1]) of the original array, and find a contiguous subarray giving maximum profit. If the difference falls below 0, reset it to zero.

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

*maxCur = current maximum value

*maxSoFar = maximum value found so far

maximum subarray也可以用这个解