[LeetCode] 121. Best Time to Buy and Sell Stock

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本文介绍了一种解决股票买卖最佳时机问题的算法。通过一次遍历记录最低价格并计算最大利润，适用于仅允许进行一次交易的情况。同时提供两种实现思路及代码示例。

题：https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/

#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 (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

有两种，第一种更好：

1.

从头遍历，如果记录最低值，并计算扫过点与最低点差值的最大值。

2.

首先，从后向前遍历一遍，获取一个dict，记录了从dict[i] 为从i以后天数最大的 maxprice 。
然后，从头遍历，用当前price与dict[i]比较，获取最大的profit

#code

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if len(prices)==0:
            return 0
        
        minprice = prices[0]
        maxprofit = 0
        for i in range(1,len(prices)):
            if minprice>prices[i]:
                minprice = prices[i]
            elif maxprofit<prices[i] - minprice:
                maxprofit = prices[i] - minprice
        return maxprofit

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        
        maxsellprice  = {}
        tmpmax = 0
        for i in range(len(prices)-1,0,-1):
            if tmpmax<prices[i]:
                tmpmax = prices[i]
            maxsellprice[i] = tmpmax
        tmpmaxprofit = 0
        for i in range(len(prices)-1):
            if tmpmaxprofit< maxsellprice[i+1] - prices[i]:
                tmpmaxprofit = maxsellprice[i+1] - prices[i]
        return tmpmaxprofit

第二版

只要记录遍历处前面的最小价格，将这个最小价格作为买入价格，然后将当前的价格作为售出价格，查看当前收益是不是最大收益。

class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length == 0)
            return 0;
        int curMin = prices[0];
        int MaxRes = 0;
        for(int i = 1 ; i < prices.length; i++){
            if(prices[i]<curMin)
                curMin = prices[i];
            else
                MaxRes = Math.max(MaxRes,prices[i] - curMin);
        }
        return MaxRes;
    }
}