Leetcode121. Best Time to Buy and Sell Stock-优快云博客

本文链接：https://blog.youkuaiyun.com/mohuak/article/details/77976106

本文介绍了一个经典的算法问题——寻找股票买卖的最佳时机以获得最大利润。通过一次遍历数组，使用贪心算法找到最大差价，即为最大利润。

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Leetcode121. Best Time to Buy and Sell Stock

题目：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.

题目分析：题目要求我们只能交易一次而取得最大的利润，由此来看，题目并不是很难，只需要利用贪心算法，遍历一次就可以得到最终的结果，因此这道题也很容易就得到了结果。

代码：

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