題目描述：

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.


同樣是動態規劃的題目，但這次是求元素間最大的差，且必須是後面的元素減去前面元素所得的最大差（具有潛在的時間屬性），這裏其實一個用遍歷就能解決

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

14.Best Time to Buy and Sell Stock