LeetCode(121) Best Time to Buy and Sell Stock
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题目
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
求买卖最大收益问题。简单来说就是给定一个整数序列,求最大差值。
该题目的关键是,效率。
我们很容易想到二次循环解决该问题,但是很不幸,是超时的! 所以,必须另择他法。
其实在O(n)时间内,也可以解决问题,只需要加一个记录当前最低价格的变量即可;价低购入,价高售出嘛~
AC代码
class Solution {public: int maxProfit(vector<int>& prices) { if (prices.empty()) return 0; //maximum记录最大收益,price记录当前最低价格 int maximum = 0, price = prices[0] ,size = prices.size(); for (int i = 1; i < size ; ++i) { //价低时买入 if (prices[i] < price) { price = prices[i]; continue; } //价高时卖出,比较并记录最大收益 else{ int tmp = prices[i] - price; if (tmp > maximum) maximum = tmp; } }//for return maximum; }};
GitHub测试程序源码
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