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.

解题思路：

　　之前一道的简单模式，想法有点相似。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size()==0||prices.size()==1) return 0;
        int max_profit=0;
        int min_buyer =prices[0];

        for(int i=1;i<prices.size();i++){

         if(prices[i]-min_buyer>max_profit) max_profit=   prices[i]-min_buyer;
         if(prices[i]<min_buyer) min_buyer = prices[i];

        }
        return max_profit;
    }
};

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