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Best Time to Buy and Sell Stock III

2024-07-05 10:47:20   224人阅读

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 (ie, you must sell the stock before you buy again).

思路：最佳时间买入卖出股票，但是这题与前两题有所不同而且比较难，最多进行两次买卖，但是同一时间手上只能保持一个股票。使用动态规划思想，是比较好的思路(PS:也是借鉴网上大神思路).先正序扫描0~i天的最大利润，在倒序扫描i+1~n天的最大利润，放在两个数组中，然后求出第i项最大和。

1.min1=prices[0];

2.定义profit1数组，保存正序扫描的利润：profit1[i]=profit1[i]-min(prices[i],min1)当profit1[i]-min(prices[i],min1)>profit1[i-1];反之则为profit1[i]=proft1[i-1];

3.max1=prices[n-1]

4.定义profit2数组，保存逆序扫描的最大利润:profit2[i]=max(prices[i],max1)-prices[i],当max(prices[i],max1)-prices[i]>profit2[i+1]；反之，profit2[i]=profit2[i+1];

5.最后把第i项最大和求出，返回最大利润。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int n=prices.size();
        if(n<=0)
            return 0;
        vector<int> profit1(n,0);
        vector<int> profit2(n,0);
        int min=prices[0];
        for(int i=1;i<n-1;i++)
        {
            if(prices[i]<min)
                min=prices[i];
            if(prices[i]-min>profit1[i-1])
                profit1[i]=prices[i]-min;
            else
                profit1[i]=profit1[i-1];
        }
        int max=prices[n-1];
        for(int i=n-2;i>=0;i--)
        {
            if(prices[i]>max)
                max=prices[i];
            if(max-prices[i]>profit2[i+1])
                profit2[i]=max-prices[i];
            else
                profit2[i]=profit2[i+1];
        }
        int maxprofits=0;
        for(int i=0;i<n-1;i++)
        {
            if(profit1[i]+profit2[i]>maxprofits)
                maxprofits=profit1[i]+profit2[i];
        }
        return maxprofits;
    }
};

 


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