LeetCode Best Time to Buy and Sell Stock III class Solution { public: int maxProfit(vector<int> &prices) { int len = prices.size(); if (len < 2) return 0; vector<int> i2r(len, 0); vector<int> i2l(len, 0); int max_price = prices[len - 1...
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). 解答: 1. 从左往右扫描,计算0-i的这个区间的最大利润。方法可以参见股票第一题 2. 从右往左扫描,计算i-len这个区间的最大利润。方法同上。 3. 再从头至尾扫一次,每个节点加上左边...
Can you solve this real interview question? Best Time to Buy and Sell Stock III - You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete at most two transacti
123. Best Time to Buy and Sell Stock III 参考了别人的思想,大概是两重动态规划问题,一个值记录前i天profit最大值,另一个值记录i天以后最大值,两者合一即为最大买买次数为2的时候的最大值。代码如下:...123. Best Time to Buy and Sell Stock III 这道题用自己的笨方法超时,想到用dp但是不知道...
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. Example 2: Input: [1,2,3,4,5]
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. Example 2: Input: [1,2,3,4,5]
Best Time to Buy and Sell Stock 题目大意 给定每天的股票价格,如果只允许进行一轮交易,也就是买进一次和卖出一次,求所能获得的最大的利润。 解题思路 DP或者直接解决 代码 DP dp[i] = max(dp[i - 1], prices[i] - minPrice) minPrice永远是之前的最底价格 ...
A transaction is a buy & a sell. You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). Analysis Comparing to I and II, III limits the number of transactions to 2. This can be solve by “devide and conquer”. We use left...
Design an algorithm to find the maximum profit. You may complete at mosttwotransactions. Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). 分析: 依照题意,要求依据每天的股票价格。交易两次得到最大的利润。直接的想法就是利用...
- 3.买入卖出一次股票,即完成了一次交易sell1 - 4.买入卖出一次股票, 并且还买入一次股票buy2 - 5.买入卖出两次股票,即完成了两次交易sell2 不进行任何操作的收益为0,可以不用考虑。那么只需要考虑剩下的4个状态。 buy1=max{buy1‘,−prices[i]}, 可以有前一天的buy1或者当天...