标签：
123 Best Time to Buy and Sell Stock III
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).
//http://yucoding.blogspot.com/2012/12/leetcode-question-10-best-time-to-buy.html
//http://www.cnblogs.com/springfor/p/3877068.html
就是分成两段一段是(0-i)一次买卖,一段是(i+1 -- len-1)一次买卖。上述两次买卖的最大利益。。
动态规划，左数组保存，0-i的时候一次交易的最大收益。(从0就是最左边开始到某个位置i：一次买卖最大收益)
右数组保存，i-最后位置的时候，就是i到len-1时候的最大收益。(从某个位置i到最后的位置len-1：一次买卖的最大收益)
然后找到i点的最大收益：就是从i点之前的一次交易加上i点之后的一次交易的最大收益（按照题目要求的两次买卖）。
1 public class bestbuysell123 { 2 public int maxProfit(int[] prices){ 3 if(prices.length==0 || prices==null){ 4 return 0; 5 } 6 7 int len = prices.length; 8 int[] left = new int[len]; 9 int[] right = new int[len]; 10 11 left[0] = 0; 12 int low = prices[0]; 13 for(int i = 1; i<len; i++){ 14 low = Math.min(low, prices[i]); 15 left[i] = Math.max(left[i-1], prices[i]-low); 16 } 17 18 right[0]=0; 19 int hight = prices[len-1]; 20 for(int i =len-2; i>=0; i--){ 21 hight = Math.max(hight, prices[i]); 22 right[i] = Math.max(right[i+1], hight-prices[i]); 23 } 24 25 int maxfit = 0; 26 for(int i = 0; i < len; i++){ 27 maxfit=Math.max(maxfit, left[i]+right[i]); 28 } 29 return maxfit; 30 } 31 }
想了半天，想明白了就好了。
LeetCode 123 Best Time to Buy and Sell Stock III
标签：
原文地址：http://www.cnblogs.com/hewx/p/4535698.html
