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).

public class Solution {
    /**
	 * To solve this problem, we use two array to save the max profit.
	 * The running time is O(n)
	 * 
	 * @param prices --input price array
	 * @return maximum profit 
	 * @author Averill Zheng
	 * @version 2014-06-04
	 * @since JDK 1.7
	 */
	public int maxProfit(int[] prices) {
		int maxProfit = 0;
		int length = prices.length;
		if(length > 0){
			//maximum in order
			int[] maxInOrder = new int[length];
			maxInOrder[0] = 0; 
			int min = prices[0];
			for(int i = 1; i < length; ++i){
				if(prices[i] - min >= maxInOrder[i - 1])
					maxInOrder[i] = prices[i] - min;
				else
					maxInOrder[i] = maxInOrder[i - 1];
				min = (prices[i] < min) ? prices[i] : min;
			}
			//maximum in reverse order
			int[] maxInReverseOrder = new int[length];
			maxInReverseOrder[length - 1] =0;
			int max = prices[length - 1];
			for(int i = length -2; i > -1; --i){
				if(max - prices[i] > maxInReverseOrder[i + 1])
					maxInReverseOrder[i] = max - prices[i];
				else
					maxInReverseOrder[i] = maxInReverseOrder[i + 1];
				max = (prices[i] > max) ? prices[i] : max;
			}
			//find the maxProfit 
			for(int k = 0; k < length; ++k)
				maxProfit = (maxProfit < maxInOrder[k] + maxInReverseOrder[k]) ? maxInOrder[k] + maxInReverseOrder[k] : maxProfit;
		}
		return maxProfit;
    }  
}