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.
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.
一刷
題解：維護(hù)一個price的當(dāng)前最小值刻撒， 然后對于每個price, 用與最小值的差值update最大的profit
public class Solution {
public int maxProfit(int[] prices) {
if(prices == null || prices.length == 0) return 0;
int maxProfit = 0, minPrice = Integer.MAX_VALUE;
for(int price : prices){
if(price<minPrice) minPrice = price;
maxProfit = Math.max(maxProfit, price-minPrice);
}
return maxProfit;
}
}
