[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).

https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/

 

思路：因为两次交易不能重叠，所以需要扫描两边分别求两段的最大值。dp[i]代表prices[0..i]的最大值，dpRev[i]代表prices[i..n]的最大值。然后求出max(dp[i]+dpRev[i])  0<=i<=n-1

public class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length < 2)
            return 0;
        int n = prices.length;
        int res = 0;
        int[] dp = new int[n];
        int[] dpRev = new int[n];
        dp[0] = 0;
        dpRev[n - 1] = 0;
        int curMin = prices[0];

        for (int i = 1; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], prices[i] - curMin);
            if (prices[i] < curMin)
                curMin = prices[i];
        }
        int curMax = prices[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            dpRev[i] = Math.max(dpRev[i + 1], curMax - prices[i]);
            if (prices[i] > curMax)
                curMax = prices[i];

            res = Math.max(res, dp[i] + dpRev[i]);
        }

        return res;
    }
}

 

第二遍记录：

注意dp 和 dpDev代表的具体含义。

 

第三遍记录：

　　注意当天卖了之后可以再买

　　注意写的时候小心，prices,dp,dpRev不要写错了。 

 

参考：

http://www.cnblogs.com/TenosDoIt/p/3436457.html