LeetCode刷题——Array篇(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.

1、保存买入价价，卖出时比较收益是否大于原收益，大则更新收益，同时，如果出现更低的买入价，则更新买入价；
2、有没有可能更新买入价后，出现后面的售价和之前的买入价相比更大。答案是否定，因为如果出现更低的话肯定是和现在的更低价进行比较；

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size()<2)
            return 0;
        int sell = prices[0];
        int result = 0;
        for (int i = 1; i < prices.size(); i++)
        {
            if (prices[i] - sell > result)
                result = prices[i] - sell;
            if (prices[i] < sell)
                sell = prices[i];
        }
        return result;

    }
};