标签：des style blog color os io for 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.
没有算法的思维真可怕，n^2肯定超时，但还是写了。最后参考了别人的思路，用了动态规划，应该算是一道典型的动态规划题。
思路如下：
设dp[i]是[0,1,2...i]区间的最大利润，则该问题的一维动态规划方程如下
dp[i+1] = max{dp[i], prices[i+1] - minprices} ,minprices是区间[0,1,2...,i]内的最低价格
1 class Solution{ 2 public: 3 int maxProfit(vector<int> &prices){ 4 int len=prices.size(); 5 if(len<=1) 6 { 7 return 0; 8 } 9 int max=prices[1]-prices[0],minprice=prices[0]; 10 for(int i=2;i<len;i++) 11 { 12 minprice=prices[i-1]<minprice?prices[i-1]:minprice; 13 if(max<prices[i]-minprice) 14 { 15 max=prices[i]-minprice; 16 } 17 } 18 if(max<0) 19 { 20 return 0; 21 } 22 else 23 { 24 return max; 25 } 26 } 27 };
LeetCode:Best Time to Buy and Sell Stock,布布扣,bubuko.com
LeetCode:Best Time to Buy and Sell Stock
标签：des style blog color os io for 问题
原文地址：http://www.cnblogs.com/levicode/p/3869836.html
