Leetcode Best Time to Buy and Sell Stock

题目：

Say you have an array for which the i^th 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.

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数组代表每天股票的价格，只能买入卖出一次，求最大利润。

本质上，就是对数组求最大差值。遍历数组，通过求当前最小值与比其大的数的差值（利润），并记录最大差值（利润），一边遍历，一边更新。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
    	if(prices == NULL || prices.size() == 0) {
    		return 0;
    	}
    	
    	int min = prices[0];
    	int pro = 0;
    	for(int i = 1; i < prices.size(); i++) {
    		if(min > prices[i]) {
    			min = prices[i];
    		}
    		else {
    			if(pro < prices[i] - min) {
    				pro = prices[i] - min;
    			}
    		}
    	}
    	return pro;
    }
};