leetcode-121. 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.

方法一思路：（好想但是复杂度高）
一个数组表示股票每天价格，求买入卖出最大收益
看每天的价格假设买入，然后在看后面每天价格假设卖出求出收益，一天天的遍历，如果比原来的收益大就更新。
主要就是注意一下第二个for循环是从i开始（即从今天往后的日子才能卖）。
代码：

class Solution {public:    int maxProfit(vector<int>& prices) {        int maxProfit=0;//最大收益        for(int i=0;i<prices.size();++i){            for(int j=i;j<prices.size();++j){                if((prices[j]-prices[i])>=maxProfit){//更新最大收益                    maxProfit=prices[j]-prices[i];                }            }        }        return maxProfit;    }};

方法二思路：（比一好一点）
先设定一个购买价格为prices[0]，如果后面比它小，肯定不能卖，把购买价格更新；如果后面比它大，就假设卖出，把maxProfit更新。如果后面在比它大在重新更新maxProfit，比它小就从新更新buyPrice，这样相当于设置了两个指针，maxProfit总是最大的。

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.size()==0)            return 0;        int maxProfit=0;        int buyPrice=prices[0];        int tempProfit=0;        for(int i=0;i<prices.size();++i){            if(prices[i]<buyPrice){                buyPrice=prices[i];            }            tempProfit=prices[i]-buyPrice;            if(tempProfit>maxProfit){                maxProfit=tempProfit;            }        }        return maxProfit;    }};