buy and sell stocks 在leetcode现在总共有5 道题。 Best Time to Buy and Sell Stock 这道题是基础,只能transaction一次,所以我们用 Best Time to Buy and Sell Stock II 这道题表示可以一直transaction,所以只需要下一次比上一次高,那就加一次。 Best Time to Buy and Sell Stock with Cooldown 这道题属...
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). However, you may not engage in ...
2) Either I didn't hold stocks yesterday, but today I chose to buy, so today I hold stocks. This explanation should be clear. If you buy, you need to subtract prices[i] from the profit, and if you sell, you need to increase prices[i] to the profit. The maximum profit of ...
local(n, k) = max(local(n-1, k)+diff, global(n-1, k-1), global(n-1,k-1)+diff); 三种情况: 1,第n-1天卖出。而且第n-1天买入,然后第n天卖出。那尽管是k+1次交易,可是最后两次交易能够合并成一次,因此也算是k次交易 2,前n-1天内已经进行了k-1次交易。然后第n天买入,并卖出 3。前...
1.题目描述 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 (ie, you must sell the ...
res+=tempreturnres 参考: 下面的更快,因为索引查找的次数少一些! classSolution(object):defmaxProfit(self, prices):""":type prices: List[int] :rtype: int"""iflen(prices) ==0:return0else: profit=0 start=prices[0]foriinrange(1,len(prices)):ifprices[i]>start: ...
Best time to buy and sell stocks IV 题目 https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/ 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....
View Code 别人代码: View Code 学习之处: diff的变量命名不错 if(null==stocks || stocks.length==0) return 0; int maxProfit = 0 ; int cur = stocks[0]; for(int stock : stocks){ if(stock > cur) maxProfit += (stock-cur);