0. 题目

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.

1. c++版本

方法1，使用暴力法，此方法不能AC，算法复杂度O(n_2)

    int maxProfit(vector<int>& prices) {
        int maxProfit = 0;
        for (int i=0; i<prices.size(); ++i)
            for (int j=i+1; j <prices.size(); ++j) {
                if(prices[i] <= prices[j]) {
                    int currentProfit = prices[j] - prices[i];
                    maxProfit = max(currentProfit, maxProfit);
                }
            }
        return maxProfit;
    }

方法2：只允许一次交易，思路就是去找到在前一段区间内最小的数，和后一段区间内最大的数，二者之差即为最大收益。

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

2. python版本

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices:
            return 0
        minValue, maxPro = prices[0], 0
        for i in range(1, len(prices)):
            minValue = minValue if minValue < prices[i] else prices[i]
            tmpValue = prices[i] - minValue
            maxPro = maxPro if maxPro > tmpValue else tmpValue
        return maxPro;