LeetCode-123. Best Time to Buy and Sell Stock III

LeetCode-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).

Example 1:

Input: [3,3,5,0,0,3,1,4]Output: 6Explanation: 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.

Example 2:

Input: [1,2,3,4,5]Output: 4Explanation: 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.

Example 3:

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

题解：

看别人的。。

一种n次购买的很精巧的计算方法：

buy[i]是当前余额在i时买入之后最大所剩额度，sell[i]是在i时卖掉之后最大利润，多次遍历。

class Solution {public: int maxProfitWithNTrans(vector& prices, int n) { int l = prices.size(); if (l < 2) { return 0; } vector buy(n, 0), sell(n, 0); for (int i = 0; i < n; i++) { buy[i] = -prices[0]; } for (int i = 1; i < l; i++) { for (int j = 0; j < n; j++) { if (j == 0) { buy[j] = max(buy[j], -prices[i]); } else { buy[j] = max(buy[j], -prices[i] + sell[j - 1]); } sell[j] = max(sell[j], prices[i] + buy[j]); } } return sell[n - 1]; } int maxProfit(vector& prices) { return maxProfitWithNTrans(prices, 2); }};

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