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最新推荐文章于 2024-01-05 16:26:30 发布

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最新推荐文章于 2024-01-05 16:26:30 发布

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本文链接：https://blog.youkuaiyun.com/tuzhq3/article/details/83902641

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本文探讨了如何找到股票买卖的最佳时机以获取最大利润。通过两种算法思路，分别采用两重循环和维护最小值与最大利润的方法，实现了从股票价格数组中找出最佳买入和卖出时机的问题。第一种方法虽然直观但效率较低，第二种方法则更加高效，只遍历一次数组即可得到答案。

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121. Best Time to Buy and Sell Stock

Description：

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [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.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

想要求出最大利益，有两种解题思路。

思路一：

用两重循环遍历所有可能的交易，求出最大利益，时间复杂度为O(n^2)。

代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size() == 0) return 0;
        int maxProfit = 0;
        for(int i = 0; i < prices.size(); i++) {
            for (int j = i+1; j < prices.size(); j++) {
                if (prices[j]-prices[i] > maxProfit)
                    maxProfit = prices[j]-prices[i];
            }
        }
        return maxProfit;
    }  
};

思路二：

维护一个当前最大利益maxProfit以及最小值minPrice，只需遍历一次即可求出最大利益，时间复杂度为O(n)。

代码如下：

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