问题
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.
思路
前面总是找到最小值，然后后面每走一步就计算一次与该最小值之间的差，把该差与预设的最大值作比较。
代码1 顺利通过
class Solution { public: int maxProfit(vector<int>& prices) { if(prices.empty()) { return 0; } int max=0; int min=prices[0]; int len=prices.size(); for(int i=0;i<len;i++) { if(prices[i]<min) { min=prices[i]; } if(prices[i]-min>max) { max=prices[i]-min; } } return max; } };
代码2：常规思路 效率O（n*n） 显示超时
class Solution { public: int maxProfit(vector<int>& prices) { int max=0; int len=prices.size(); for(int i=0;i<len;i++) { for(int j=i+1;j<len;j++) { int temp=0; if(prices[j]>prices[i]) { temp=prices[j]-prices[i]; } if(temp>max) { max=temp; } } } return max; } };
@121. Best Time to Buy and Sell Stock----Array----Easy---20160926
原文：http://www.cnblogs.com/lingyi1111/p/5909696.html