标签：des style blog io ar color os 使用 sp
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.
首先，问题转换为数学语言：对数组A[n], 对任意i < j < n, 求出max(A[j] - A[i]).
可以使用O(n^2)的方法得到结果。
以下分析动态规划方法：
对于输入prices[n], 如果已知profit[i - 1],
1) 当prices[i] <= prices[i - 1]时，显然，profit[i] = profit[i - 1]
2)当prices[i] > prices[i - 1]时, 记bottom = min(prices[0, 1, ..., i]), 如果
prices[i] - bottom > profit[i - 1] 则有
profit[i] = prices[i] - bottom
否则，profit[i] = profit[i - 1]
1 int maxProfit(vector<int> &prices) 2 { 3 int n = prices.size(), bottom, max_profit; 4 if (n <= 1) 5 return 0; 6 int *profit = new int[n]; 7 profit[0] = 0; 8 bottom = prices[0]; 9 for (int i = 1; i < n; i++) 10 { 11 bottom = min(bottom, prices[i]); 12 if (prices[i] > prices[i - 1] && prices[i] - bottom > profit[i - 1]) 13 { 14 profit[i] = prices[i] - bottom; 15 } 16 else 17 { 18 profit[i] = profit[i - 1]; 19 } 20 } 21 max_profit = profit[n - 1]; 22 delete profit; 23 return max_profit; 24 }
leetcode. Best Time to Buy and Sell Stock
标签：des style blog io ar color os 使用 sp
原文地址：http://www.cnblogs.com/ym65536/p/4158613.html
