Leetcode 121. best time to Buy and Sell Stock

121. Best time to Buy and Sell Stock

• Total accepted:115636
• Total submissions:313848
• Difficulty:easy

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 1:

Input: [71536456- 1 5 7-16 as selling price needs to being larger than buying price)

Example 2:

Input: [764310this case  is0.

Idea: DP.

Method One: For the first day, find [0,i-1] Day buy the lowest price, thus get the first day of sale of the highest price. Then traverse I to get the highest selling price.

Method Two: Maximal sub-sequences and problems.

Code:

Method One:

1 classSolution {2  Public:3     intMaxprofit (vector<int>&prices) {4         intN=prices.size (), i,maxprofit=0, minprice=Int_max;5          for(i=0; i<n;i++){6             if(maxprofit<prices[i]-minprice) {7maxprofit=prices[i]-Minprice;8             }9             if(minprice>Prices[i]) {TenMinprice=Prices[i]; One             } A         } -         returnMaxprofit; -     } the};

Method Two:

Reference:

https://discuss.leetcode.com/topic/19853/ Kadane-s-algorithm-since-no-one-has-mentioned-about-this-so-far-in-case-if-interviewer-twists-the-input

Https://en.wikipedia.org/wiki/Maximum_subarray_problem

Original:

The logic to solve ThisProblem isSame as "Max Subarray Problem" usingKadane's algorithm. Since no body have mentioned this so far, I thought it's a good thing foreverybody to know. All the straight forward solution shouldifThe interviewer twists the question slightly by giving the difference array of prices, Ex: for{1,7,4, One},ifHe gives {0,6, -3,7}, you might end up being confused. Here, the logic isTo calculate the difference (maxcur + = prices[i]-prices[i-1]) of the original array, and find a contiguous subarray giving maximum profit. If the difference falls below0, reset it to zero.  Public intMaxprofit (int[] prices) {        intMaxcur =0, Maxsofar =0;  for(inti =1; i < prices.length; i++) {Maxcur= Math.max (0, maxcur + = prices[i]-prices[i-1]); Maxsofar=Math.max (Maxcur, Maxsofar); }        returnMaxsofar; }*maxcur =Current Maximum value*maxsofar = maximum value found so far

1 classSolution {2  Public:3     intMaxprofit (vector<int>&prices) {4         intI,n=prices.size (), curprofit=0, maxprofit=0;5          for(i=1; i<n;i++){6Curprofit=max (0, curprofit+prices[i]-prices[i-1]);7maxprofit=Max (maxprofit,curprofit);8         }9         returnMaxprofit;Ten     } One};

Leetcode 121. best time to Buy and Sell Stock