LeetCode—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.

其实就是在一个数组当中找到最大的差值，较大的值在小值后面

常规的想法就是一个二维的循环，从当前值开始向后遍历一遍获得最大值，但是这样的时间复杂度是o(n2)

其实主要是因为要求最大值在后面，所以其实可以从后面向前面进行遍历，记录插值最大，而且一直记录最大值

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

或者从前面开始记录最小值也是可以的：

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