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.
分析一：使用两个辅助数组min_prices和max_prices。min_prices的第i个元素表示到第i天前最小的价格。max_prices的第i个元素表示第i天后最大的价格。时间复杂度O(n),空间复杂度O(n)
class Solution { public: int maxProfit(vector<int>& prices) {
if(prices.empty()) return 0; int prices_size = prices.size(); vector<int> min_prices(prices_size, prices[0]); vector<int> max_prices(prices_size, prices[prices_size - 1]); for (int i = 1; i < prices_size; i++) { if (prices[i] < min_prices[i-1]) { min_prices[i] = prices[i]; } else { min_prices[i] = min_prices[i-1]; } } for (int i = prices_size - 2; i >= 0 ; i--) { if (prices[i] > max_prices[i+1]) { max_prices[i] = prices[i]; } else { max_prices[i] = max_prices[i+1]; } } int max_gap = 0; for (int i = 0; i < prices_size; i++) { int temp_gap = max_prices[i] - min_prices[i]; if (temp_gap > max_gap) max_gap = temp_gap; } return max_gap; } };
分析二：我们完全可以不借助于这两个数组。时间复杂度O(n),空间复杂度O(1)
class Solution { public: int maxProfit(vector<int>& prices) { if (prices.empty()) return 0; int min_price = prices[0]; int max_profit = 0; for (int i = 1; i < prices.size(); i++) { if (prices[i] < min_price) { min_price = prices[i]; } max_profit = max(max_profit, prices[i] - min_price); } return max_profit; } };
[LeetCode] Best Time to Buy and Sell Stock
原文：http://www.cnblogs.com/vincently/p/4844766.html