leetcode 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).

public class S123 {
    //分成两段求最大子数组,
    //forward[i]表示1~i这段最大子数组和，backward[i]表示从prices.length-1~1这段最大子数组和
    //最后找出forward[i]与backward[i+1]和最大值即所求
    //求forward[i]与backward[i]的方法与053kadane's algorithm类似
    public int maxProfit(int[] prices) {
        if(prices.length<2)
            return 0;
        int result =0;
        int sum = 0;
        int []forward = new int[prices.length];
        forward[0] = Integer.MIN_VALUE;
        int []backward = new int[prices.length+1];
        backward[prices.length] = Integer.MIN_VALUE;
        for(int i = prices.length-1;i>0;i--){
            prices[i] = prices[i]-prices[i-1];
            if(sum<0) sum = 0;
            sum+=prices[i];
            backward[i] = sum>backward[i+1]?sum:backward[i+1];
        }
        sum = 0;
        for(int i = 1;i<prices.length;i++){
            if(sum<0) sum = 0;
            sum+=prices[i];
            forward[i] = sum>forward[i-1]?sum:forward[i-1];
            if(forward[i]>0&&backward[i+1]>0){
                result = forward[i]+backward[i+1]>result?forward[i]+backward[i+1]:result;
            }else{
                result = forward[i]>result?forward[i]:result;
                result = backward[i+1]>result?backward[i+1]:result;
            }
        }
        return result;
    }
}