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second sale of the selling point and selling points are likely to be maxind.Realize:Case1 is easy, is to run two times the best trading, merit to join. If there is no valid trade, it is recorded as 0.The best Buy and sell before Minind is recorded as PARTGAP2 after Partgap1,maxind.Case1result = Maxgap + Max (PARTGAP1, PARTGAP2);Case2 more trouble, let us explain in detail:Simple analysis reveals:The value of two transactions is Where Minleft is the minimum value before Minind, including the Min

Here are three questions with maximum subarray, you can look at this question firstI)Say you has an array for which the i-th element is the price of a given-stock on day I.If you were-permitted-to-complete at most one transaction (ie, buy one and sell one share of the stock), design an AL Gorithm to find the maximum profit.The first, dynamic programming algorithm for this problem. State transfer equation F (i) = Max (Maxprofit, Prices[i]-minprice)

121.Say you has an array for which the i-th element is the price of a given-stock on day I.If you were-permitted-to-complete at most one transaction (ie, buy one and sell one share of the stock), design an AL Gorithm to find the maximum profit.classSolution { Public: intMaxprofit (vectorint>prices) { intn =prices.size (); if(N 1) return 0; intMini = prices[0], ans =0, I; for(i =1; I )

Topic:The pigs can fly under the tuyere. The bull market in China today is a "missed seven years". Give you an opportunity to review the history of a stock, known for a continuous N-day price movement, in the length of an integer array of n, the array of elements I (Prices[i]) represents the stock price of the first day. Suppose you don't have a stock at first, but you have the opportunity to buy 1 shares and then sell 1 shares at most two times, and

classSolution { Public intMaxprofit (int[] prices) { intMaxprofit = 0; intSecondmaxprofit = 0; intProfit = 0; intThistimemaxprofit = 0; intBuyday = 0; intMaxbuyday = 0; intBuyvalue; if(Prices.length = = 0){ return0; }Else{Buyvalue= Prices[0]; Buyday= 0; } for(inti = 0; i ) {Profit= Prices[i]-Buyvalue; if(Profit >Maxprofit) {Maxp

1. Best time to Buy and Sell Stock ITopic linksTitle Requirements:Say you has an array for which the i-th element is the price of a given-stock on day I.If you were-permitted-to-complete at most one transaction (ie, buy one and sell one share of the stock), design an AL Gorithm to find the maximum profit.The essence of this problem is to find out the maximum difference between any two numbers in an array (the number of large numbers minus the small numbers). We find that the maximum difference i

Best time to Buy and Sell Stock ISay you has an array for which the i-th element is the price of a given-stock on day I.If you were-permitted-to-complete at most one transaction (ie, buy one and sell one share of the stock), design an AL Gorithm to find the maximum profit.ExampleGiven array [3,2,3,1,2] , return 1 .Analysis: Because the sell is always after buying, so as long as there is a lower buy price, we can update the buy price, if the price is lower than the buy price, we will update Tempm

Say you has an array for which the i-th element is the price of a given-stock on day I.Design an algorithm to find the maximum profit. You are in most of the transactions.Ideas:1, the topic requests most trades two times, if only trades once, the question is "best time to Buy and Sell the Stock";If the transaction must be traded two times, the first transaction is sold on the I trading day, then the problem degrades into the maximum benefit of 0 to I trading day and i+1 to the last trading day.

title :Say you has an array for which the i-th element is the price of a given-stock on day I.Design an algorithm to find the maximum profit. You could complete as many transactions as (ie, buy one and sell one share of the stock multiple times). However, engage in multiple transactions for the same time (ie, you must sell the stock before you buy again).code : OJ Test via runtime:71 ms1 classSolution:2 #@param prices, a list of integers3 #@re

Update Leetcode to solve Java answers on a regular basis.Choose from the pick one approach.As a result of the random series of problems, continue to do later versions.Test instructions is given an array, the array represents the price of the item of the day, allowing multiple transactions (the premise of selling is to buy, buy the premise is to sell, that is, the body has no items to buy), ask the maximum benefit is how much.The first way to analyze this is when we should buy and when we sell.Im

price[0..j+1] can be introduced at O (1) from PRICE[0..J].But how to get from Price[j. N-1] launched Price[j+1..n-1]? In turn thinking, we can use O (1) time by price[j+1..n-1] to launch price[j. N-1].The final algorithm:Array L[i] records the maximum profit of PRICE[0..I],Array R[i] records the price[i. N] the maximum profit.Known L[i], seeking l[i+1] is simple, also known r[i], R[i-1] is also very easy.Finally, we use O (n) time to find the largest l[i]+r[i], which is the problem. (The final

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The pigs can fly under the tuyere. The bull market in China today is a "missed seven years". Give you an opportunity to review the history of a stock, known for a continuous N-day price movement, in the length of an integer array of n, the array of elements I (Prices[i]) represents the stock price of the first day. Suppose you don't have a stock at first, but you have the opportunity to buy 1 shares and then sell 1 shares at most two times, and make s

[] prices) { if(Prices = =NULL|| prices.length==0 | | Prices.length ==1)return0; int[] Leftprofit =New int[Prices.length]; intMinprice = Prices[0]; intMaxprofit = 0; for(inti=1;i//Find the maximum profit that can be obtained from the front and back, the maximum profit for the current price minus the previous minimum price and the previous maximum profit o