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.
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这道题一开始没看出来是动态规划。。。看出来就好做了。。所以要注意识别。
public class Solution { public int maxProfit(int[] prices) { if (prices == null || prices.length == 0) { return 0; } int[] res = new int[prices.length]; int low = prices[0]; res[0] = 0; for (int i = 1; i < prices.length; i++) { if (prices[i] < low) { low = prices[i]; } res[i] = Math.max(res[i - 1], prices[i] - low); } return res[prices.length - 1]; } }
当然这个代码还可以优化一下，优化成O(1)的空间
public class Solution { public int maxProfit(int[] prices) { if (prices == null || prices.length == 0) { return 0; } int max = 0; int low = prices[0]; for (int i = 1; i < prices.length; i++) { if (prices[i] > low) { max = Math.max(max, prices[i] - low); } else { low = prices[i]; } } return max; } }
注意一下第二种方法的条件判断。
Best Time to Buy and Sell Stock Leetcode
原文：http://www.cnblogs.com/aprilyang/p/6298672.html