121. Best Time to Buy and Sell Stock

121. Best Time to Buy and Sell Stock

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 (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.

代码如下：

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

解题思路：

注意，此处的minPrice是到当天为止，价格最小的股票价格，并不是所有价格中最低的，直到计算到最后一天。同理maxPro计算的是到当前为止，当前股票价格与到目前为止的最小值的差值，差值越大利润越大。