Leetcode:121. Best Time to Buy and Sell Stock

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 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.
本題關鍵是要先買後賣，而且買的價格一定要比賣的少，否則輸出爲0

class Solution {
    public:
    int maxProfit(vector<int>& prices) {  
        if (prices.size() == 0) return 0;
        int max = 0, min = prices[0];
        int profit = 0;
        for (int i = 1; i < prices.size(); i++) {
            if (prices[i] < min) {
                min = prices[i];
            } else {
                if (prices[i] - min > profit) {
                    profit = prices[i] - min;
                }
            }
        }
     return profit;
    }
};