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).

題意：121的followup，現(xiàn)在要求只能進(jìn)行兩次交易，求最大獲利。

思路：
自己沒有想出解法，參考了discuss的解法。
discuss采用動態(tài)規(guī)劃來解這道題，dp[i][j]代表在第j天交易i次能獲得的最大利潤。
如果在第j天沒有賣出行為，則第j天最大獲利與第j-1天最大獲利相等，即dp[i][j]等于dp[i][j-1]。
如果在第j天進(jìn)行賣出，那么最后一次買入在第1天到第j-1天都可能發(fā)生，所以dp[i][j]=prices[j] - prices[jj] + dp[i-][jj] { jj in range of [0, j-1] }。
綜合以上兩種情況：
dp[i][j] = max(dp[i][j-1], prices[j] - prices[jj] + dp[i-1][jj] { jj in range of [0, j-1] })
prices[j]是定值，可以用tmpMax來表示dp[i-1][jj] - prices[jj]{ jj in range of [0, j-1] }.

public int maxProfit(int[] prices) {
    if (prices == null || prices.length < 2) {
        return 0;
    }

    int profit = 0;
    int k = 2;
    int len = prices.length;
    int[][] dp = new int[k][len];
    for (int i = 1; i < k; i++) {
        int tmpMax = dp[i-1][0] - prices[0];
        for (int j = 1; j < len; j++) {
            dp[i][j] = prices[j] + tmpMax;
            tmpMax = Math.max(tmpMax, dp[i-1][j] - prices[j]);
            profit = Math.max(profit, dp[i][j]);
        }
    }

    return profit;
}