Java for LeetCode 123 Best Time to Buy and Sell Stock III【HARD】

Say you have an array for which the i^th 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).

解题思路一：

既然是两次交易的话，分为左右两个区间即可。先按照一次交易的思路算出交易的最大值，存一个数组里，然后从后往前遍历，找到符合条件的和最大的两个，JAVA实现如下：

	public int maxProfit(int[] prices) {
		if (prices.length == 0)
			return 0;
		int[] oneProfit = new int[prices.length];
		int buy_price = prices[0], profit = 0;
		for (int i = 1; i < prices.length; i++) {
			buy_price = Math.min(buy_price, prices[i]);
			profit = Math.max(profit, prices[i] - buy_price);
			oneProfit[i] = profit;
		}
		int res = oneProfit[prices.length - 1];
		int sell_price = prices[prices.length - 1];
		profit = 0;
		for (int i = prices.length - 1; i >= 1; i--) {
			sell_price = Math.max(sell_price, prices[i]);
			profit = Math.max(profit, sell_price - prices[i]);
			res = Math.max(res, profit + oneProfit[i - 1]);
		}
		return res;
	}

解题思路二：

一种类似提前购买的思路：

参考https://leetcode.com/discuss/18330/is-it-best-solution-with-o-n-o-1%E3%80%82

JAVA实现如下：

    public int maxProfit(int[] prices) {
        int hold1 = Integer.MIN_VALUE, hold2 = Integer.MIN_VALUE;
        int release1 = 0, release2 = 0;
        for(int i:prices){                              // Assume we only have 0 money at first
            release2 = Math.max(release2, hold2+i);     // The maximum if we've just sold 2nd stock so far.
            hold2    = Math.max(hold2,    release1-i);  // The maximum if we've just buy  2nd stock so far.
            release1 = Math.max(release1, hold1+i);     // The maximum if we've just sold 1nd stock so far.
            hold1    = Math.max(hold1,    -i);          // The maximum if we've just buy  1st stock so far.
        }
        return release2; ///Since release1 is initiated as 0, so release2 will always higher than release1.
    }