121. Best Time to Buy and Sell Stock（补上周）-优快云博客

本文链接：https://blog.youkuaiyun.com/fuyetao/article/details/72862759

本文介绍了一种寻找股票买卖最佳时机以实现最大利润的算法。该算法只需遍历一次价格数组，通过记录最低购买价格并计算当前利润是否为最大值来确定最佳买卖时机。

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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.

题目分析：这是卖股票的第一个题目，根据题意我们知道只能进行一次交易，但需要获得最大的利润，

所以我们需要在最低价买入，最高价卖出，当然买入一定要在卖出之前。

对于这一题，不算那么难，只需要遍历一次数组，通过一个变量记录当前最低价格，同时算出此

次交易利润，并与当前最大值比较就可以了。

代码如下：

class Solution {
public:
int maxProfit(vector<int> &prices) {
if(prices.size() <= 1) {
return 0;
}
int minP = prices[0];
int profit = prices[1] - prices[0];
for(int i = 2; i < prices.size(); i++) {
minP = min(prices[i - 1], minP);
profit = max(profit, prices[i] - minP);
}
if(profit < 0) {
return 0;
}
return profit;
}
};