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: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.
思路：
比较典型的动态规划，实时更新当前出现过的最小价格，并将目前遍历到的价格与最小值比较，如果大于目前的最高收益就将最大收益更新。
解法：
1 public class Solution 2 { 3 public int maxProfit(int[] prices) 4 { 5 if(prices.length < 2) 6 return 0; 7 8 int maxValue = 0; 9 int minPrice = prices[0];10 11 for(int i = 0; i < prices.length; i++)12 {13 if(prices[i] < minPrice)14 minPrice = prices[i];15 if(prices[i] - minPrice > maxValue)16 maxValue = prices[i] - minPrice;17 }18 19 return maxValue;20 }21 }