标签:art vector inline tran submit ati .net info tar
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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.
题意:给定一组数,表示某种股票每天的交易价格。求怎样进行一次买卖。使收益最大。
思路:贪心
买进价越低越好,卖出价越高越好。终于的目的是利润越高越好
设置两个变量,一个表示当前的最低卖出价cur_min_price,还有一个表示当前的最高利润max_profit
遍历数组prices。以当前值为卖出价
更新
max_profit = max(max_profit, prices[i] - cur_min_price)
cur_min_price = min(cur_min_price, prices[i])
复杂度:时间O(n),空间O(1)
相关题目:
Best Time to Buy and Sell Stock II
Best Time to Buy and Sell Stock III
class Solution { public: int maxProfit(vector<int> &prices) { if (prices.size() < 2) return 0; int cur_min_price = prices[0]; int max_profit = 0; for(int i = 1; i < prices.size(); i++){ max_profit = max(max_profit, prices[i] - cur_min_price); cur_min_price = min(cur_min_price, prices[i]); } return max_profit; } };
Leetcode 贪心 Best Time to Buy and Sell Stock
标签:art vector inline tran submit ati .net info tar
原文地址:http://www.cnblogs.com/lytwajue/p/7089289.html