Best Time to Buy and Sell Stock

LeetCode #121

Description:

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 sell one share of the stock), design an algorithm to find the maximum profit.

Example:

Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.

Idea:

跟maximum sum subarray类似，但是更简单。每步计算目前最大的profit，与max_profit对比。然后更新之前的min price。

Time： O(n).

Code:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty()) return 0;

        int max_profit = 0;
        int min_price=prices[0];

        for(int i=1; i<prices.size(); ++i){
            if(prices[i]-min_price > max_profit)
                max_profit = prices[i]-min_price;
            if(prices[i]<min_price)
                min_price=prices[i];
        }

        return max_profit;
    }
};