Best Time to Buy and Sell Stock III

Question

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 (i.e., you must sell the stock before you buy again).

Example 1:
Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3

LeetCodeleetcode.com/problems/best-time-to-buy-and-sell-stock-iii

Solution

/*
brute force的优化
双向dp: 一个从左往右计算max profit, 另一个从右往左计算，最后分割，然后取max
*/
public int maxProfit(int[] prices) {
    if (prices.length <= 1) {
        return 0;
    }
    int len = prices.length;
    int[] left = new int[len];
    int[] right = new int[len];
    
    int min = prices[0];
    left[0] = 0;        
    for (int i = 1; i < prices.length; i++) {                        
        if (prices[i] > prices[i-1]) {
            left[i] = Math.max(prices[i]-min, left[i-1]);
        }else {
            left[i] = left[i-1];
            min = Math.min(min, prices[i]);
        }
    }
        
    
    //注意计算从右往左的时候是记录max
    int max = prices[len-1];
    right[len-1] = 0;        
    for (int i = len-2; i >= 0; i--) {                                    
        if (prices[i] < prices[i+1]) {
            right[i] = Math.max(max-prices[i], right[i+1]);
        }else {
            right[i] = right[i+1];
            max = Math.max(max, prices[i]);
        }
    }
    
    int res = Math.max(left[len-1], right[0]);
    for (int i = 1; i < len-1; i++) {
        res = Math.max(left[i]+right[i+1], res); 
    }        
    return res;
}