LeetCode OJ 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).

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【题目分析】

用一个数组表示股票每天的价格，数组的第i个数表示股票在第i天的价格。最多交易两次，手上最多只能持有一支股票，求最大收益。

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【思路】

动态规划法。以第i天为分界线，计算第i天之前进行一次交易的最大收益preProfit[i]，和第i天之后进行一次交易的最大收益postProfit[i]，最后遍历一遍，max{preProfit[i] + postProfit[i]} (0≤i≤n-1)就是最大收益。第i天之前和第i天之后进行一次的最大收益求法同Best Time to Buy and Sell Stock I。

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【java代码】

 public class Solution {
     public int maxProfit(int[] prices) {
         if(prices.length < 2) return 0;

         int n = prices.length;
         int preProfit[] = new int[n];
         int postProfit[] = new int[n];

         int curMin = prices[0];
         for(int i = 1; i < n; i++){
             curMin = Math.min(curMin, prices[i]);
             preProfit[i] = Math.max(preProfit[i-1], prices[i] - curMin);
         }

         int curMax = prices[n-1];
         for(int i = n-2; i >= 0; i--){
             curMax = Math.max(curMax, prices[i]);
             postProfit[i] = Math.max(postProfit[i+1], curMax - prices[i]);
         }

         int maxProfit = 0;
         for (int i = 0; i < n; i++) {
             maxProfit = Math.max(maxProfit, preProfit[i] + postProfit[i]);
         }

         return  maxProfit;
     }
 }