buy foxconn stock

Source of the topic:https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/ Test Instructions Analysis:Similar to the above question, Array[i] represents the price of items for the first day, if only 2 times can be traded. Ask for maximum profit. Topic Ideas:This is a dynamic planning issue. It is not difficult to think of splitting the entire array into two parts. Then use two arrays

Question Say you have an array for which the ith element is the price of a given stock on day I. 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. Answer To maximize the benefits, buy at the lowest price and se

Title Link: click~/* Test instructions: An array, the element I represents the price of the stock of day I, only allowed to buy and sell once, for maximum profit *//** * Ideas: Two pointers, one scan the entire array, the other record the minimum value of the scanned element * During the scan, update the minimum value while updating the maximum profit * Maximum profit: Max (A[j]-a[i]), J>i */class

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 encrypt the stock before you buy again ).1: special case; 2: traverse two numbers from front to back and f

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. The idea without algorithms is terrible. N ^ 2 definitely times out, but it is still written. Finally, I took a look at oth

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. Class Solution { Public : Int Maxprofit (vector Int > Prices ){ Int Profit = 0 , Minprice; Int Len = Prices. Siz

Three types of stock trading algorithmsFirst, there is no limit to the number of transactionsUse greedy strategy to find the longest increment sequence and accumulate the corresponding profit.Second, only one transactionUsing the dynamic programming algorithm, the minimum price before the corresponding time node is recorded from the point of travel, and the maximum profit at this node is obtained while the minimum price is updated.Three, up to two tim

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

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