should buy square stock

Original title address: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/The content is in the code and comments, it is not wordy.Class Solution {Public:int Maxprofit (vectorbest time to Buy and Sell Stock--leetcode

Say you have an array for whichITh element is the price of a given stock on dayI. If you were only permitted to complete at most one transaction (ie, buy one and every one share of the stock), design an algorithm to find the maximum profit. At the beginning, I thought it was the maximum value minus the minimum value. However, because the purchase occurred before

Title Link: click~/* Test instructions: An array, the element I represents the price of the stock of day I, allow up to two times to buy and sell, for maximum profit *//** * idea: Record the maximum profit of the day with the Currprofit array, scan the array from start to finish to get the * Currprofit = max (Currpor Fit[i], prices[i]-low) * * Record the maximum profit after the day with a futureprofit arra

Idea: Add two of the largest profit, the maximum profit of a period is not necessarily only in one rise period, for example, there are two ascending intervals between the Green Line As long as the maximum profit before I is calculated and the maximum profit after I is added, because I cannot be a peak or a valley value at the same time, it is guaranteed not to buy and sell on the same day The Green Line profit plus the red line profit is the biggest

The topics are as follows: best time to buy and Sell the total accepted:43912 total submissions:135635 My submissions question Solution Say you have an array for which the ith element are the price of a given the If you are were only permitted to complete in most one transaction (ie, buy one and sell one share of the "stock", design an AL Gorithm to find the max

Problem description: Say you have an array for whichITh element is the price of a given stock on dayI. If you were only permitted to complete at most one transaction (ie, buy one and every one share of the stock), design an algorithm to find the maximum profit. Solution: 1 Public int maxprofit (INT [] prices) {2 if (prices. length

Last night and classmates to do the game, randomly picked the problem. My idea.The first thing that comes to mind is that the first number is traversed, and then a maximum number is found from the back of the number, which is, of course, the time complexity of the Loop loop O (N2).The method is feasible, but time does not pass when the data becomes very large. And then thinkFor example, there is already a benefit, then how to produce a bigger profit, the factor is that my input price is lower, o

Update Leetcode to solve Java answers on a regular basis.Choose from the pick one approach.Test instructions to allow only one sale, the given array is the price of the item of the day, ask the maximum benefit is how much.Obviously the best way to get the most benefit is to iterate through the array, and in the process of traversal, the price of each day is read, and it is compared with the minimum value obtained by the preceding traversal, which is the maximum benefit to sell a single time on t

Public classS121 { Public intMaxprofit (int[] prices) { //TLE/*if (prices.length*/ //divide and conquer Method--from the introduction of algorithm maximum Subarray problem, AC but slow, move findcrossmax () inside Findmax () makes 3ms quicker// not Best/* if (PRICES.LENGTH*/ // Best One if(prices.length) return0; intCurmin = Prices[0]; intRET =Integer.min_value; for(inti = 1;i) {curmin=math.min (Curmin, prices[i]); RET= Math.max (ret, prices[i]-curm

Personally, this is a relatively difficult DP problem. This link posts a typical DP solution to it. Need some time to get how it works.The code is rewritten as follows.1 classSolution {2 Public:3 intMaxprofit (vectorint>prices) {4 intn = prices.size (), num =2;5 if(N 1)return 0;6vectorint> > dp (num +1, vectorint> (n,0));7 for(intK =1; K ) {8 inttemp = dp[k-1][0]-prices[0];9 for(inti =1; I ) {TenDp[k][i] = max (Dp[k][i-1], Prices[i] +temp);

Algorithm:Scan through the array, keep finding1. Prices[i] with the condition:prices[i] 2. Prices[i] with the condition:prices[i] > prices[i+1], or at the end of a ascending trend, treat it as selling PO IntKeeping till reach the end of the array;　　Note:1. If a buying point is recorded and does reach the end, then must exist a selling point after it (must exist some point L Arger than buying point)2. If reach the end when finding the buying point, then stop and finish the whole programe.best tim

It is easy to know that you are seeking a continuous maximum number of sub-arrays. But note the boundary conditions, assuming the sum of the maximum subarray of 0, then do not trade, return 0.public class Solution {public int maxprofit (int[] prices) { if (Prices.length Leetcode-best time to Buy and Sell Stock

Traverse the price vector, add up the profit for each period of the ascent, and remember to calculate the last profit at the end. Class Solution { //Add up profit for each period of ascent //Minval: Trough value //maxval: Peak //Profit: Profit public: int Maxprofit ( Vector profit + = Maxval-minval; Minval = Maxval = Prices[i]; } If up, change the peak else{ maxval = prices[i]; } } Finall

* * * * File:stock_price.cpp * Author:hongbin * gives a stock price sequence to find the best buy and sell point, that is, after the sequence of elements with the maximum value of the preceding elements. * #include

: The profit earned at the current price (if you don't make money, you don't sell).①, if Curmax = 0, means that the current value is smaller than the value of the front;②, Curmax > 0, i.e., Curmax = (price[i-1]-price[i-2] + price[i]-price[i-1]) = Price[i]-price[i-1]. That is, the current price is the maximum value, and the lowest value is subtracted from the front.Max: The maximum profit in the total process. Public int maxProfit2 (int[] prices) { int curmax = 0, max = 0;

Class Solution {public: int Maxprofit (vectorint Maxprofit (vectorBest solution, feel good slag.As long as there is an increase can be added to the sum inside ... There is no need to calculate the rising range.Still have to analyze the problem carefully.[Leetcode] best time to Buy and Sell Stock II