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LeetCode123:Best Time to Buy and Sell Stock III

2024-07-06 04:49:28   220人阅读

题目：

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).

解题思路：

话说这题同前两题难度瞬间就拉开好多，哎，编程能力还是不行啊，如果不是谷歌各路大神解题报告http://blog.csdn.net/pickless/article/details/12034365，真心想不出来。

这题实际上用到了DP和分段的思想。

首先，根据题意，要求至少买卖两次（就因为有这限制，使得题目难度突然就增加了），所以，我们可以进行分段。

寻找一个点i，将原来的price[0..n-1]分割为price[0..i]和price[i..n-1]，分别求两段的最大profit，可知分段就是使得买卖至少进行两次。

下面求price[0..i]和price[i..n-1]两段的最大profit时，利用了DP思想。

对于点i+1，求price[0..i+1]的最大profit时，很多工作是重复的，在求price[0..i]的最大profit中已经做过了。

类似于Best Time to Buy and Sell Stock，可以在O(1)的时间从price[0..i]推出price[0..i+1]的最大profit。

但是如何从price[i..n-1]推出price[i+1..n-1]？反过来思考，我们可以用O(1)的时间由price[i+1..n-1]推出price[i..n-1]。

最终算法：

数组l[i]记录了price[0..i]的最大profit，

数组r[i]记录了price[i..n]的最大profit。

已知l[i]，求l[i+1]是简单的，同样已知r[i]，求r[i-1]也很容易。

最后，我们再用O(n)的时间找出最大的l[i]+r[i]，即为题目所求。

实现代码：

#include <iostream>
#include <vector>
using namespace std;

/**
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).


*/

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        if(prices.empty())
            return 0;
        int n = prices.size();
        int *l = new int[n];
        int *r = new int[n];
        l[0] = 0;
        int lmin = prices[0];
        for(int i = 1; i < n; i++)
        {
            lmin = min(prices[i],lmin);
            l[i] = max(l[i-1], prices[i] - lmin);
        }
        
        r[n-1] = 0;
        int rmax = prices[n-1];
        for(int i = n - 2; i >= 0; i--)
        {
            rmax = max(rmax, prices[i]);
            r[i] = max(r[i+1], rmax - prices[i]);
        }
        
        int maxprofit = 0;
        for(int i = 0; i < n; i++)
        {
            maxprofit = max(maxprofit, l[i] + r[i]);
        }
        delete l;
        delete r;
        return maxprofit;
        
    }
};

int main(void)
{
    int arr[] = {2,4,5,1,7,10};
    int n = sizeof(arr) / sizeof(arr[0]);
    vector<int> stock(arr, arr+n);
    Solution solution;
    int max = solution.maxProfit(stock);
    cout<<max<<endl;
    return 0;
}


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