LeetCode | 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).

要求：最多进行两次transaction，且同时只能持有一只股票。 tags：dynamic programming

解法一：

基本的想法：将数组分为前、后两个部分，分别用Best Time to Buy and Sell Stock I的方法来求解两个部分的最大收益，再求和。
通过N次数组划分之后，选出最大的总利润（提示超时。。。。。）

public class Solution {    public int profit(int[] prices, int start, int end){  //Best Time to Buy and Sell Stock I 的解法        if(end <= start) return 0;          int maxProfit = 0;          int min_price = prices[start];          for(int i=start; i<=end; i++){              if(prices[i]<min_price) min_price=prices[i];                  if(prices[i]-min_price > maxProfit) maxProfit=prices[i]-min_price;          }          return maxProfit;      }        public int maxProfit(int[] prices) {       //leetcode函数，实现对数组的N次划分        if(prices.length <= 1) return 0;                int maxProfit = 0;        for(int i=0; i<prices.length; i++){            int sum_profit = profit(prices, 0, i) + profit(prices, i+1, prices.length-1);            if(maxProfit < sum_profit) maxProfit = sum_profit;        }        return maxProfit;    }}

解法二：

profit_0_i表示从0至i天所能获得的最大利润
profit_i_n表示从i至n天所能获得的最大利润
举例：prices = {2 1 8 3 5 9}，则：
  profit_0_i = {0 0 7 7 7 8}，填充时从左向右遍历，总是取已知最低价格min_price，然后用prices[i]-min_price来判断[0-i]区间能获得的最大收益
  profit_i_n = {8 8 6 6 4 0}，填充时从右向左遍历，总是取已知最高价格max_price，然后用max_price-prices[i]来判断[i-n]区间能获得的最大收益
 则 max{ profit_0_i[i] + profit_i_n[i] } = 7 + 6 = 13 即为最大收益

public class Solution {    public int maxProfit(int[] prices) {               if(prices.length <= 1) return 0;                int[] profit_0_i = new int[prices.length];     //记录从0至i能获得的最大收益        int[] profit_i_n = new int[prices.length];     //记录从i至n能获得的最大收益                //求解填充数组profit_0_i        int temp_1 = 0;        int min_price = prices[0];        for(int i=0; i<prices.length; i++){       //同Best Time to Buy and Sell Stock I 中的方法            if(prices[i] < min_price) min_price = prices[i];            if(prices[i]-min_price > temp_1) temp_1 = prices[i] - min_price;            profit_0_i[i] = temp_1;        }                //求解填充数组profit_i_n        int temp_2 = 0;        int max_price = prices[prices.length-1];        for(int j=prices.length-1; j>=0; j--){   //同Best Time to Buy and Sell Stock I 中的方法            if(prices[j] > max_price) max_price = prices[j];            if(max_price-prices[j] > temp_2) temp_2 = max_price - prices[j];            profit_i_n[j] = temp_2;        }                 //找最大的收益和        int maxProfit = 0;        for(int k=0; k<prices.length; k++){       //找最大的对应项之和            int temp = profit_0_i[k] + profit_i_n[k];            if(temp > maxProfit) maxProfit = temp;        }        return maxProfit;    }}