Leetcode 121. best time to Buy and Sell stock when stock sale (dynamic planning, array, simulation)

Title Description

An array is known, and the I element represents the price of the stock of day I, you can only trade once (buy and sell each time), design algorithms to find the maximum benefit

Test examples

Input: [7, 1, 5, 3, 6, 4]Output: 5最大收益 = 6-1 = 5 (不是7-1 = 6,因为先买后卖，7买，1买亏了6)Input: [7, 6, 4, 3, 1]Output: 0最大收益为0

Detailed analysis

At first glance, it's very simple to iterate through an array, select an element at a time, find the maximum value of the array following the element, calculate the difference, and compare the current maximum benefit, like this:
[7, 1,5,3,6,4] current 7, rear Max 6, yield 1
[7,1, 5,3,6,4] current 1, rear max 6, yield 5
[7,1,5, 3,6,4] current 5, rear Max 6, yield 1
So continue to find.
However, this method will time out, a little change, now not only to maintain the maximum benefits, but also to save the lowest price, if the price is lower today, the minimum price (Buy), and if the price of the day minus the minimum price of the largest return, the maximum price (sell), the process is as follows:

[7, 1,5,3,6,4] minPrice=INT_MAX，maxProfit=0 =minPrice=7,maxProfit=0

[7,1, 5,3,6,4] minPrice=7，maxProfit=0 =minPrice=1,maxProfit=0

[7,1,5, 3,6,4] minPrice=1，maxProfit=0 =minPrice=1,maxProfit=4

[7,1,5,3, 6,4] minPrice=1，maxProfit=4 =minPrice=1,maxProfit=4

[7,1,5,3,6, 4] minPrice=1，maxProfit=4 =minPrice=1,maxProfit=5

[7,1,5,3,6,4] minPrice=1，maxProfit=5 =minPrice=1,maxProfit=5

Algorithm implementation

• Method 1:time Limit exceeded

   Solution {public : int  maxprofit (vector<    int  >& prices) {int  Profit = 0 ; for  (auto  start=prices.begin (); Start!=prices.end (); start++)        {int  buy = *start;        int  sell = *std::max_element (Start,prices.end ()); if  ((sell-buy) >profit)        {Profit = Sell-buy; }} return  profit;}};  

• Method 2:accepted

  class  Solution{public : int  maxprofit (vector<int     > &prices) {int  Profit = 0 ;    int  minbuy = Std::numeric_limits<int  >::max (); for  (int  i=0 ; i<prices.size (); i++) {//buy it if current price was less than minimum prices yet  if (prices[i]<minbuy) {minbuy = prices[i]; } //sell it if current profit got optimally  if  ((prices[i]-mi        Nbuy) >profit) {profit = Prices[i]-minbuy; }} return  profit;}};  

Leetcode 121. best time to Buy and Sell stock (dynamic planning, array, simulation) when selling the stock of the year (dynamically planned)