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.

只进行一次交易，使得到最大的利润。首先想到使用蛮力算法两层for循环遍历数组。结果导致超时。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0 || prices.size() == 1)
            return 0;
        int res = 0, mindiff = INT_MAX;
        for (int i = 0; i != prices.size() - 1; i++) {
            for (int j = i + 1; j != prices.size(); j++) {
                int tmp = prices[j] - prices[i];
                if (tmp >= 0)
                    res = max(res, tmp);
            }
        }
        return res;
    }
};

TLE

接着想了一个办法就是遍历一次数组。在遍历的同时比较出最小的购买价格，然后用这一天以后的每天价格减去这个最小购买价格取最大值即可。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0 || prices.size() == 1)
            return 0;
        int res = 0, minDiff = INT_MAX;
        for (int i = 0; i != prices.size(); i++) {
            minDiff = min(minDiff, prices[i]);
            res = max(res, prices[i] - minDiff);
        }
        return res;
    }
};
// 6 ms