leetcode 123. Best Time to Buy and Sell Stock III

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最新推荐文章于 2025-04-18 22:24:21 发布

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本文介绍了一种算法，用于寻找给定股票价格序列中进行两次交易以获得最大利润的策略。通过实例说明了如何在不同时段买入和卖出股票以最大化收益。

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123. Best Time to Buy and Sell Stock III

Hard

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Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

class Solution {
public:
    int maxProfit(vector<int>& prices) 
    {    
        int n = prices.size();
        if(n<2) return 0;
        vector<int> maxr = maxrange(prices);
        vector<int> maxrev = maxrangerev(prices);
        int maxpro = maxr[n-1];
        for(int i=0;i<n-1;i++)
        {
            maxpro = max(maxpro,maxr[i]+maxrev[i+1]);
        }
        return maxpro;
    }
    vector<int> maxrange(vector<int>& price)
    {
        int n = price.size();
        vector<int> maxr(n,0);
        int minl=price[0];
        for(int right=1;right<n;right++)
        {
            maxr[right] = max(maxr[right-1],price[right]-minl);
            if(price[right]<minl) minl = price[right];
        }
        return maxr;       
    }
    vector<int> maxrangerev(vector<int>& price)
    {
        int n = price.size();
        vector<int> maxr(n,0);
        int maxright = price[n-1];
        for(int left=n-2;left>=0;left--)
        {
            maxr[left] = max(maxr[left+1],maxright-price[left]);
            if(price[left]>maxright) maxright = price[left];
        }
        return maxr;
    }
};