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

本文链接：https://blog.youkuaiyun.com/ghscarecrow/article/details/86620711

本文介绍了一种算法，用于计算给定股票价格数组中能够获得的最大利润，仅允许进行一次买卖交易。通过寻找最低购买价格并计算后续可能的利润，最终得出最大利润。

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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 (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.

思路：由于收益 = 卖出价格 - 买入价格

所以我们只需找到最小价格的下标（天数），再通过迭代不断更新maxprofit的值。由于卖出不能在买入之前，所以最小价格minprice应该先判断先更新

具体解题代码如下：

class Solution {
    public int maxProfit(int[] prices) {
        int maxprofit = 0;
        int minprice = Integer.MAX_VALUE;
        if(prices.length <= 0 || prices == null) {
            return maxprofit;
        }
        for(int i = 0;i < prices.length;i++) {
            if(prices[i] < minprice) {
                minprice = prices[i];
            } else if(prices[i] - minprice > maxprofit) {
                maxprofit = prices[i] - minprice;
            }
        }
        return maxprofit;
    }
}