[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).

Solution:

从左至右，从右至左，分别扫一遍，得到在该天以前，以后分别进行一次操作的利润   l2r[i] , r2l[i]。

 

public class Solution {
    public int maxProfit(int[] prices) {
        int N = prices.length;
        if (N < 2)
            return 0;
        int[] l2r = new int[N];
        int[] r2l = new int[N];
        int minn = prices[0];
        l2r[0] = 0;
        for (int i = 1; i < N; ++i) {
            minn = Math.min(minn, prices[i]);
            l2r[i] = Math.max(l2r[i - 1], prices[i] - minn);
        }
        r2l[N - 1] = 0;
        int maxx = prices[N - 1];
        for(int i=N-2;i>=0;--i){
            maxx=Math.max(maxx, prices[i]);
            r2l[i]=Math.max(r2l[i+1], maxx-prices[i]);
        }
        int result=l2r[N-1];
        for(int i=0;i<N-1;i++){
            result=Math.max(result, l2r[i]+r2l[i+1]);
        }
        return result;
    }
}

需要注意的是：

1： 存在只进行一次操作，记得到最优解的可能（题目：You may complete at most two transactions.）。要对result赋初值。

int result=l2r[N-1];

2： 你不能当天进行两次操作，今天卖出了，明天才能买进。

result=Math.max(result, l2r[i]+r2l[i+1]);