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This article is in the study summary, welcome reprint but please specify Source: http://blog.csdn.net/pistolove/article/details/43024967Say 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 maxi

real-time monitoring of NetEase stock market competition players buy shares details . First Look at the software used and the interface to be monitored:650) this.width=650; "Width=" "class=" Zoom "id=" aimg_141 "src=" http://www.mutousoft.cn/bbs/data/attachment/ Forum/201508/26/223644sy442c4lkk4ym5gg.jpg "alt=" 223644sy442c4lkk4ym5gg.jpg "/> 650" this.width=650; "Widt H= "class=" Zoom "id=" aimg_142 "src="

Problem Description: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.Note:Engage in multiple transactions on the same time (ie, you must sell the stock before you buy again).Analysis:According to test instructions. Ask for the maximum profit

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

Leetcode: Best Time to Buy and keep Stock IV, leetcode= 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 k transactions.Note:You may not engage in multiple transactions at the same time (ie, you must sell the stock