LeetCode --Best Time to Buy and Sell Stock III

LeetCode –Best Time to Buy and Sell Stock III

技术与社会生态的相互作用，使得技术发展经常带来更深远的问题，远远超出了技术设备的直接目的和运作本身，同样的技术，引入不同的环境，或在不同条件下，会带来非常不同的结果

---

题目描述

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).

至多交易2次，求最大利益。

算法解释

1. basically we want to maximize the formula: -price1 + price2 - price3 + price4, where every price appear in that order. So every time we visit a new price, we update the max possible value of 4 (partial) formula. That’s why we use max for each of the 4 states.
   也就是状态机4个状态。和cooldown相似。

在每个步骤中，我们可以看到，oneBuy始终是输入数组中的最低价格，oneBuyOneSell跟踪价格和最低价格之间的最大差异，TwoBuy的价值变化反映了输入价格数组中的谷底。
变量TwoBuyTwoSell保持直到目前的价格的最大的利润，
（代码比较清晰易懂，请看代码）

1. DP

f[k, ii] represents the max profit up until prices[ii] (Note: NOT ending with prices[ii]) using at most k transactions.

---

f[k, ii] = max(f[k, ii-1], prices[ii] - prices[jj] + f[k-1, jj]) { jj in range of [0, ii-1] }
= max(f[k, ii-1], prices[ii] + max(f[k-1, jj] - prices[jj]))

      f[0, ii] = 0; 0 times transation makes 0 profit
      f[k, 0] = 0; if there is only one price data point you can't make any money no matter how many times you can trade

状态转移方程明了了，解释略

代码

public int maxProfit(int[] prices) {
    int oneBuy = Integer.MIN_VALUE;
    int oneBuyOneSell = 0;
    int twoBuy = Integer.MIN_VALUE;
    int twoBuyTwoSell = 0;
    for(int i = 0; i < prices.length; i++){
        oneBuy = Math.max(oneBuy, -prices[i]);//we set prices to negative, so the calculation of profit will be convenient
        oneBuyOneSell = Math.max(oneBuyOneSell, prices[i] + oneBuy);
        twoBuy = Math.max(twoBuy, oneBuyOneSell - prices[i]);//we can buy the second only after first is sold
        twoBuyTwoSell = Math.max(twoBuyTwoSell, twoBuy + prices[i]);
    }

    return Math.max(oneBuyOneSell, twoBuyTwoSell);
}

使用DP的方法

class Solution {
public:
    int maxProfit(vector<int> &prices) {

        if (prices.size() <= 1) return 0;
        else {
            int K = 2; // number of max transation allowed
            int maxProf = 0;
            vector<vector<int>> f(K+1, vector<int>(prices.size(), 0));
            for (int kk = 1; kk <= K; kk++) {
                int tmpMax = f[kk-1][0] - prices[0];
                for (int ii = 1; ii < prices.size(); ii++) {
                    f[kk][ii] = max(f[kk][ii-1], prices[ii] + tmpMax);
                    tmpMax = max(tmpMax, f[kk-1][ii] - prices[ii]);
                    maxProf = max(f[kk][ii], maxProf);
                }
            }
            return maxProf;
        }
    }
};