题目

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).

思路

• 一开始是想定义一个点i，d[i]表示在i点之前卖掉第一次股票所得到的最大利润，这个狂野的想法很明显是要TLE的，最关键的问题是，进行了过多的重复计算，[i…n]和[i+1…n]的最大利润在prices[i+1] < prices[i]的时候是一样的

• 参考了别人的思路，分别算[0…i]的最大利润和[i…n]的最大利润，很明显[0…i]的最大利润的算法跟之前Stock1的解法是一样的，但是后面[i…n]的解法就更为暴力一点，从后往前扫，遇到最高的就记下来，最大利润就是当前max(highest - prices[i], maxproofit)。

代码

public class Solution {
    public int maxProfit(int[] prices) {

        if (prices.length == 0) {
            return 0;
        }

        // calculating [0...i] max profit
        int[] leftMaxProfit = new int[prices.length];

        // calculating [i...n] max profit
        int[] rightMaxProfit = new int[prices.length];


        leftMaxProfit[0] = 0;
        for (int i = 1; i < leftMaxProfit.length; i++) {
            int changes = prices[i] - prices[i - 1];
            leftMaxProfit[i] = leftMaxProfit[i - 1] + changes < 0 ? 0 : leftMaxProfit[i - 1] + changes;
        }

        rightMaxProfit[0] = 0;
        int highest = prices[prices.length - 1];
        int maxprofit = Integer.MIN_VALUE;
        int ret = 0;
        for (int i = rightMaxProfit.length - 1; i >= 1; i--) {
            int profit = highest - prices[i];
            if (profit > maxprofit) maxprofit = profit;
            int finalprofit = maxprofit + leftMaxProfit[i];
            if (finalprofit > ret) ret = finalprofit;
            if (prices[i] > highest) highest = prices[i];
        }
        return ret;
    }
}

Best Time To Sell Stock 3