leetcode 122. Best Time to Buy and Sell Stock II 最大子段和 + DP

蔚落 2022-06-08 04:50 72阅读 0赞

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

这道题和上一道题不一样的是，不限制次数，那么更简单了，只需要把所有的盈利都加起来。

建议和[leetcode 188. Best Time to Buy and Sell Stock IV 最大子段和][leetcode 188. Best Time to Buy and Sell Stock IV] 、[leetcode 123. Best Time to Buy and Sell Stock III 最大k次字段和 + DP][leetcode 123. Best Time to Buy and Sell Stock III _k_ _ DP] 、[leetcode 122. Best Time to Buy and Sell Stock II 最大子段和 + DP][leetcode 122. Best Time to Buy and Sell Stock II _ _ DP] 、[leetcode 121. Best Time to Buy and Sell Stock 最大字段和问题 + DP][leetcode 121. Best Time to Buy and Sell Stock _ _ DP] 、[leetcode 714. Best Time to Buy and Sell Stock with Transaction Fee 动态规划DP][leetcode 714. Best Time to Buy and Sell Stock with Transaction Fee _DP] 和[leetcode 309. Best Time to Buy and Sell Stock with Cooldown 动态规划DP][leetcode 309. Best Time to Buy and Sell Stock with Cooldown _DP]一起学习。

代码如下：

/*
     * 算法导论说这个问题可以转华为最大字段和的问题,这个题是不限制交易次数，
     * 所以遍历num数组，只要是大于0的时候，均做交易，这个题更简单
     * */
    public class Solution 
    {
        public int maxProfit(int[] prices) 
        {
            if(prices==null || prices.length<=1)
                return 0;
    
            int []num=new int[prices.length-1];
            for(int i=1;i<prices.length;i++)
                num[i-1]=prices[i]-prices[i-1];
    
            return getMaxSum(num);
        }
    
        //动态规划求解最大子段和
        int getMaxSum(int[] num)
        {
            int sum=0;
            for (int i = 0; i < num.length; i++) 
            {
                if(num[i]>0)
                    sum+=num[i];
            }
            return sum;
        }
    }

下面是C++的做法，就是遍历一次即可

代码如下：

#include <iostream>
    #include <vector>
    #include <algorithm>
    #include <climits>
    
    using namespace std;
    
    class Solution 
    {
    public:
        int maxProfit(vector<int>& p) 
        {
            if (p.size() <= 1)
                return 0;
            vector<int> dp(p.size()-1,0);
            for (int i = 0; i < dp.size(); i++)
                dp[i] = p[i + 1] - p[i];
    
            int sum = 0, b = 0;
            for (int i = 0; i < dp.size(); i++)
            {
                if (dp[i] > 0)
                    sum += dp[i];
            }
            return sum;
        }
    };

[leetcode 188. Best Time to Buy and Sell Stock IV]: http://blog.csdn.net/jackzhang_123/article/details/78037522
[leetcode 123. Best Time to Buy and Sell Stock III _k_ _ DP]: http://blog.csdn.net/jackzhang_123/article/details/77988039
[leetcode 122. Best Time to Buy and Sell Stock II _ _ DP]: http://blog.csdn.net/jackzhang_123/article/details/77987707
[leetcode 121. Best Time to Buy and Sell Stock _ _ DP]: http://blog.csdn.net/jackzhang_123/article/details/77987643
[leetcode 714. Best Time to Buy and Sell Stock with Transaction Fee _DP]: http://blog.csdn.net/jackzhang_123/article/details/78884827
[leetcode 309. Best Time to Buy and Sell Stock with Cooldown _DP]: http://blog.csdn.net/jackzhang_123/article/details/78134159