[Array] Best Time to Buy and Sell Stock

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.

方法：用简单的语言表明这道题就是指在一个数组中，用当前序号的数字减去在当前序号之前的数字，可以获得一个差值，求出在这个数组中最大的差值。可以清晰的知道被减数－减数，减数越小值会越来越大，因此求当前序号的数字差的最小值，只需要找到当前序号之前的最小数字，并减之。算法时间复杂度Ｏ(n)，空间复杂度Ｏ(1)。

class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()==0)
return 0;
int max_profit=0;
int min_number=prices[0];

for(int i=1;i<prices.size();i++){
int temp_profit = prices[i]-min_number;
if(temp_profit>max_profit)
max_profit=temp_profit;
if(prices[i]<min_number)
min_number=prices[i];
}
return max_profit;
}
};