LeetCode [121] Best Time to Buy and Sell Stock

 121. Best Time to Buy and Sell Stock

Easy

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

method1

• prices[i]-minP: the max profit if sale on the i-th day
• dp[i]: the max profit until the i-th day

method2

• the vector dp is not necessary
• only need the last value dp[i-1] when computing dp[i]
• replace the vector dp[i-1] by an integer max_profit

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//C++: method1 8ms class Solution { public: int maxProfit(vector<int>& prices) { int n = prices.size(); if(n<=1) return 0; vector<int> dp(n,0); int minP = prices[0]; for(int i=1; i<n; ++i){ minP = min(minP, prices[i]); dp[i] = max(dp[i-1], prices[i]-minP); } return dp[n-1]; } }; //C++: method1 12ms class Solution { public: int maxProfit(vector<int>& prices) { int buy = INT_MAX, maxP = 0; for(auto p:prices){ maxP = max(maxP, p-buy); buy = min(buy, p); } return maxP; } }; //C++: method2 8ms class Solution { public: int maxProfit(vector<int>& prices) { int n = prices.size(); if(n<=1) return 0; int max_profit = 0; int min_price = prices[0]; for(int i=1; i<n; ++i){ min_price = min(min_price, prices[i]); max_profit = max(max_profit, prices[i]-min_price); } return max_profit; } }; //Java class Solution { public int maxProfit(int[] prices) { int maxP = 0, buy = Integer.MAX_VALUE; for(int p : prices) { maxP = Math.max(maxP, p-buy); buy = Math.min(buy, p); } return maxP; } }