Best Time to Buy and Sell Stock(股票交易)
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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.(给定一个数组,其中第i个元素代表第i天的股价,假设最多只能进行一次交易,设计算法求最大利润)
Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5
max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.
1.个人分析
最直观的做法是通过求出数组中每个元素之间的差值来找到最大差值,但这种解法效率太低。更高效的做法是用一个变量来保存当前最小值,另一个变量保存当前最大值,在遍历数组时,将当前元素与最小值比较,并计算当前元素与最小值的差值。
2.个人解法
int maxProfit(vector<int>& prices) { if(prices.size() == 0) return 0; int profit= 0; int minVal = prices[0]; for (int i=0; i<prices.size(); ++i) { minVal = min(minVal, prices[i]); profit = max(profit, prices[i] - minVal); } return profit;}
该解法只需遍历一次数组,所以时间复杂度为O(n);这里只引进了两个变量,所以空间复杂度为O(1)。
PS:
- 题目的中文翻译是本人所作,如有偏差敬请指正。
- 其中的“个人分析”和“个人解法”均是本人最初的想法和做法,不一定是对的,只是作为一个对照和记录。
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