Best time to Buy and Sell Stock I
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.
Example
Given 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 Tempmax. Look at the code and you'll understand.
1 Public classSolution {2 /**3 * @param prices:given an integer array4 * @return: Maximum profit5 * Cnblogs.com/beiyeqingteng6 */7 Public intMaxprofit (int[] prices) {8 if(Prices = =NULL|| Prices.length <=1)return 0;9 Ten intTempmax =0; One intBuyprice = prices[0]; A - for(inti =1; i < prices.length; i++) { - if(Prices[i] >Buyprice) { theTempmax = Math.max (Tempmax, Prices[i]-buyprice); -}Else { -Buyprice =Prices[i]; - } + } - returnTempmax; + } A}
View CodeBest time to Buy and Sell Stock II
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).
Example
Given An example [2,1,2,0,1], return 2
Analysis: Since allow unlimited trading, then, each price crest is a sell point, each price trough is a buy point.
1 classSolution {2 /**3 * @param prices:given an integer array4 * @return: Maximum profit5 * cnblogs.com/beiyeqingteng/6 */7 Public intMaxprofit (int[] prices) {8 if(Prices = =NULL|| Prices.length <=1)return 0;9 Ten intBuyprice = prices[0]; One intTotal =0; A - for(inti =1; i < prices.length; i++) { - if(Prices[i] < prices[i-1]) { the if(Prices[i-1] >Buyprice) { -Total + = (Prices[i-1] -buyprice); - } -Buyprice =Prices[i]; + } - } + A if(Prices[prices.length-1] >Buyprice) { atTotal + = (Prices[prices.length-1] -buyprice); - } - returnTotal ; - } -};
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Best time to Buy and Sell Stock III
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.
Example
Given An example prices = [4,4,6,1,1,4,2,5]
, return 6
.
Analysis:
Since the title says, you can only trade at most two times, so it could be one or two times. If you are trading only once, then we will find the only one transaction maxprofit from the start point (0) to the end point (prices.length-1). If only done two times, then two times can only be produced in (0, K) and (k + 1, prices.length-1), while the range of K is 0 <= K <= prices.length-1.
1 classSolution {2 /**3 * @param prices:given an integer array4 * @return: Maximum profit5 * cnblogs.com/beiyeqingteng/6 */7 Public intMaxprofit (int[] prices) {8 if(Prices = =NULL|| Prices.length <=1)return 0;9 Ten intMax =0; One for(inti =0; i < prices.length; i++) { A if(i = = prices.length-1) { -max = Math.max (max, Maxprofit (Prices,0, i)); -}Else { themax = Math.max (max, Maxprofit (Prices,0, i) + maxprofit (prices, i +1, Prices.length-1)); - } - } - returnMax; + } - + //once one transaction is allowed from point I to J A Private intMaxprofit (int[] Prices,intIintj) { at if(I >= J)return 0; - intProfit =0; - - intLowestprice =Prices[i]; - - for(intK = i +1; K <= J; k++) { in if(Prices[k] >Lowestprice) { -Profit = Math.max (Profit, Prices[k]-lowestprice); to}Else { +Lowestprice =Prices[k]; - } the } * returnprofit; $ }Panax Notoginseng};
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best time to Buy and Sell Stock | & | | & III