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Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
This problem can be solved by "divide and conquer". We can use left[i] array to track maximum profit for transactions before i (including i), and right[i + 1] to track maximum profit for transcations after i.
Prices: 1 4 5 7 6 3 2 9 left = [0, 3, 4, 6, 6, 6, 6, 8] right= [8, 7, 7, 7, 7, 7, 7, 0]
Time complexity O(n), space cost O(n).
1 public class Solution { 2 public int maxProfit(int[] prices) { 3 if (prices == null || prices.length < 2) 4 return 0; 5 int length = prices.length, min = prices[0], max = prices[length - 1], tmpProfit = 0; 6 int[] leftProfits = new int[length]; 7 leftProfits[0] = 0; 8 int[] rightProfits = new int[length]; 9 rightProfits[length - 1] = 0; 10 // Calculat left side profits 11 for (int i = 1; i < length; i++) { 12 if (prices[i] > min) 13 tmpProfit = Math.max(tmpProfit, prices[i] - min); 14 else 15 min = prices[i]; 16 leftProfits[i] = tmpProfit; 17 } 18 // Calculate right side profits 19 tmpProfit = 0; 20 for (int j = length - 2; j >= 0; j--) { 21 if (prices[j] < max) 22 tmpProfit = Math.max(tmpProfit, max - prices[j]); 23 else 24 max = prices[j]; 25 rightProfits[j] = tmpProfit; 26 } 27 // Sum up 28 int result = Integer.MIN_VALUE; 29 for (int i = 0; i < length - 1; i++) 30 result = Math.max(result, leftProfits[i] + rightProfits[i + 1]); 31 result = Math.max(result, leftProfits[length - 1]); 32 return result; 33 } 34 }
Best Time to Buy and Sell Stock III 解答
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4825130.html
