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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.
O(1), O(1)
问题转换:即找AB对,出现次序为先A再B,使得B-A差值最大。
代码实现:一个变量记录当前出现的最小值,一个变量记录最大利润的结果。遍历数组,每次更新最小,用当前值-最小值的差去打擂台更新最大利润即可。
实现:
class Solution { public int maxProfit(int[] prices) { int min = Integer.MAX_VALUE; int ans = 0; for (int i = 0; i < prices.length; i++) { min = Math.min(min, prices[i]); ans = Math.max(ans, prices[i] - min); } return ans; } }