一、題目121：

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.

二、分析

最為簡單的賣股票問題，而且只能買賣一次，有點類似最大連續子串和題目（LeetCode53）

三、題解

class Solution {
public:
	int maxProfit(vector<int>& prices) {
		int ans = 0, tmp = 0;
		if (prices.size()<2)
			return ans;
		for (int i = 1; i < prices.size(); i++)
		{
			tmp = max(tmp + prices[i] - prices[i - 1], 0);   //類似連續子串最大和
			ans = max(ans, tmp);
		}
		return ans;
	}
};