[LeetCode122]Best Time to Buy and Sell Stock II

最新推荐文章于 2025-11-26 15:16:48 发布

转载最新推荐文章于 2025-11-26 15:16:48 发布 · 86 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zhangbaochong/p/5160174.html

文章标签：

#数据结构与算法

本文介绍了一种算法，用于计算给定每日股票价格数组时，通过多次买卖交易所能获得的最大利润。该算法采用动态规划思想，简化后的实现仅需一次遍历，比较当前价格与前一天价格，若盈利则累加至总利润。

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 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).

题意类似于股价，每天价格不一样，每天都可以出手或入手，求最终最大收益

思路:可以考虑动态规划，求出来前i-1天的最大收益，比较i天价格和i-1价格，如果大于则前i天最大收益为前i-1天最大收益加上这两天价格差，否则前i天最大收益和前i-1天最大收益相同。

代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size() == 0)
            return 0;
        vector<int> maxProfit;
        maxProfit.push_back(0);
        for(int i = 1; i < prices.size(); ++i)
        {
            if(prices[i] <= prices[i-1])
                maxProfit.push_back(maxProfit[i-1]);
            else
            {
                int tmp = prices[i] - prices[i-1];
                maxProfit.push_back(maxProfit[i-1] + tmp);
            }
        }
        int max = 0;
        for(auto num : maxProfit)
        {
            if(max < num)
                max = num;
        }
        return max;
    }
};

可以简化如下：

int maxProfit(vector<int> &prices) {
    int ret = 0;
    for (size_t p = 1; p < prices.size(); ++p) 
      ret += max(prices[p] - prices[p - 1], 0);    
    return ret;
}

转载于:https://www.cnblogs.com/zhangbaochong/p/5160174.html