※ Leetcode - Dynamic Programming - 121. Best Time to Buy and Sell Stock（动态规划）

本文介绍了一种寻找股票买卖最佳时机以获得最大利润的算法。通过两种方法实现：一种是暴力搜索，时间复杂度为O(n^2)；另一种是使用动态规划思想，将时间复杂度降低到O(n)。后者通过维护当前最小买入价格和最大利润来达到最优解。

1. Problem 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 (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.

求数组中两个索引上升的数的最大差。

2. My solution1（暴力O[n^2]）

双重循环跑一个最大差。

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

3. My solution2（DP）

每次把当前最小值记录下来，每次更新；

把当前最大值也记录下来，用当前值和最小值之差更新。

    int maxProfit(vector<int>& prices)
    {
        int minidx=0;
        int ans=0;
        int len=prices.size();
        for(int i=1; i<len; i++)
        {
            if(prices[i]<prices[minidx])
               minidx=i;
            if(prices[i]-prices[minidx]>ans)
               ans=prices[i]-prices[minidx];
        }
        return ans;
    }