64 lines
2.3 KiB
C++
64 lines
2.3 KiB
C++
// Source : https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/
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// Author : Hao Chen
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// Date : 2019-02-01
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/*****************************************************************************************************
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*
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* Say you have an array for which the ith element is the price of a given stock on day i.
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*
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* Design an algorithm to find the maximum profit. You may complete as many transactions as you like
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* (ie, buy one and sell one share of the stock multiple times) with the following restrictions:
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*
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* You may not engage in multiple transactions at the same time (ie, you must sell the stock
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* before you buy again).
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* After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)
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*
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* Example:
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*
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* Input: [1,2,3,0,2]
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* Output: 3
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* Explanation: transactions = [buy, sell, cooldown, buy, sell]
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******************************************************************************************************/
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class Solution {
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public:
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//
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//Define
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//
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// - buy[i] as the max profit when you buy the stock at day i.
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// - sell[i] as the max profit when you sell the stock at day i.
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//
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// Therefore set buy[0] = -prices[0], because spend the money, the profit is -prices[0].
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// Also set sell[0] = 0, that you do nothing in the first day.
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//
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// So,
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// buy[i] = max(buy[i-1], // do nothing - keep holding
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// sell[i-2] - prices[i] ) // sell previous day, buy today
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// // i-1 is cooldown day
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// sell[i] = max(sell[i-1], // do nothing
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// buy[i-1] + prices[i] ) // buy previous day, sell today.
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//
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int maxProfit(vector<int>& prices) {
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int len = prices.size();
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if ( len < 2 ) return 0;
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vector<int> buy(len, 0);
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vector<int> sell(len, 0);
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//day 0
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buy[0] = -prices[0];
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sell[0] = 0;
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//day 1
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buy[1] = max(buy[0], 0 - prices[1]);
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sell[1] = max(sell[0], buy[0] + prices[1]);
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for (int i=2; i<len; i++) {
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buy[i] = max( buy[i - 1], sell[i - 2] - prices[i]);
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sell[i] = max(sell[i - 1], buy[i - 1] + prices[i]);
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}
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return sell[len-1];
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}
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};
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