79 lines
2.8 KiB
C++
79 lines
2.8 KiB
C++
|
// Source : https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/
|
||
|
|
// Author : Hao Chen
|
||
|
|
// Date : 2019-02-01
|
||
|
|
|
||
|
|
/*****************************************************************************************************
|
||
|
|
*
|
||
|
|
* 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) with the following restrictions:
|
||
|
|
*
|
||
|
|
* You may not engage in multiple transactions at the same time (ie, you must sell the stock
|
||
|
|
* before you buy again).
|
||
|
|
* After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)
|
||
|
|
*
|
||
|
|
* Example:
|
||
|
|
*
|
||
|
|
* Input: [1,2,3,0,2]
|
||
|
|
* Output: 3
|
||
|
|
* Explanation: transactions = [buy, sell, cooldown, buy, sell]
|
||
|
|
******************************************************************************************************/
|
||
|
|
|
||
|
|
class Solution {
|
||
|
|
public:
|
||
|
|
//
|
||
|
|
//Define
|
||
|
|
//
|
||
|
|
// - buy[i] as the max profit when you buy the stock at day i.
|
||
|
|
// - sell[i] as the max profit when you sell the stock at day i.
|
||
|
|
//
|
||
|
|
// Therefore set buy[0] = -prices[0], because spend the money, the profit is -prices[0].
|
||
|
|
// Also set sell[0] = 0, that you do nothing in the first day.
|
||
|
|
//
|
||
|
|
// So,
|
||
|
|
//
|
||
|
|
// sell[i] = max ( buy[i-1] + prices[i],
|
||
|
|
// sell[i-1] - prices[i-1] + prices[i] );
|
||
|
|
//
|
||
|
|
// - buy[i-1]+prices[i] represents buy the stock on day i-1 and sell it on day i;
|
||
|
|
// - sell[i-1]-prices[i-1]+prices[i] represents you didn't sell the stock on day i-1
|
||
|
|
// but sell it on day i (bought stock back and sell it on day i).
|
||
|
|
//
|
||
|
|
//
|
||
|
|
// buy[i] = max ( sell[i-2] - prices[i],
|
||
|
|
// buy[i-1] + prices[i-1] - prices[i] );
|
||
|
|
//
|
||
|
|
// - sell[i-2]-prices[i] means sold the stock on day i-2 and buy it on day i
|
||
|
|
// (day i-1 is cooldown).
|
||
|
|
// - buy[i-1]+prices[i-1]-prices[i] means you didn't buy the stock on day i-1
|
||
|
|
// but buy it on day i.
|
||
|
|
//
|
||
|
|
|
||
|
|
int maxProfit(vector<int>& prices) {
|
||
|
|
int len = prices.size();
|
||
|
|
if ( len < 2 ) return 0;
|
||
|
|
vector<int> buy(len, 0);
|
||
|
|
vector<int> sell(len, 0);
|
||
|
|
|
||
|
|
buy[0] = -prices[0];
|
||
|
|
|
||
|
|
int profit = 0;
|
||
|
|
|
||
|
|
for (int i=1; i<len; i++) {
|
||
|
|
sell[i] = max ( buy[i-1] + prices[i],
|
||
|
|
sell[i-1] - prices[i-1] + prices[i] );
|
||
|
|
profit = max(profit, sell[i]);
|
||
|
|
if ( i == 1) {
|
||
|
|
buy[i] = buy[0] + prices[0] - prices[1];
|
||
|
|
} else {
|
||
|
|
buy[i] = max ( sell[i-2] - prices[i],
|
||
|
|
buy[i-1] + prices[i-1] - prices[i] );
|
||
|
|
}
|
||
|
|
|
||
|
|
}
|
||
|
|
|
||
|
|
return profit;
|
||
|
|
}
|
||
|
|
};
|