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[Leetcode]123. Best Time to Buy and Sell Stock III (C++)

题目描述

题目链接:123. Best Time to Buy and Sell Stock III

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).

例子

例子 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.

例子 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.

例子 3

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

解题思路

这道题是买卖股票的第三题,要求最多只能交易两次。参考了一下网上的做法,有一种我觉得比较好理解的方法是,维护四个变量,分别是:

  • FirstBuy : 第一次买入时的利润(应该是负值)
  • FirstSell:第一次卖出时的利润
  • SecondBuy:第二次买入时(此时第一次已卖出)的利润(可正可负)
  • SecondSell:第二次卖出时的利润

遍历一次数组,依序更新:

  • SecondSell 比较 SecondSell + prices[i]
  • SecondBuy 比较 FirstSell - prices[i] (买入,所以利润相减)
  • FirstSell 比较 FirstBuy + prices[i]
  • FirstBuy 比较 -prices[i]

代码如下:

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#include <vector>
#include<bits/stdc++.h>

class Solution {
public:
int maxProfit(std::vector<int>& prices) {
if (prices.size() <= 1) return 0;
int firstBuy = INT_MIN; //profit after first buy in, should be negative
int firstSell = 0; // profit after fisrt sell, should be positive
int secondBuy = INT_MIN; //profit after second buy in, might be negative or positive
int secondSell = 0; // profit after second sell, final answer

for (int currentPrice: prices) {
secondSell = std::max(secondSell, currentPrice + secondBuy);
secondBuy = std::max(secondBuy, firstSell - currentPrice);
firstSell = std::max(firstSell, currentPrice + firstBuy);
firstBuy = std::max(firstBuy, -currentPrice);

return secondSell < 0 ? 0 : secondSell;

}
};
  • 时间复杂度:O(n)
  • 空间复杂度:O(1)