【leetcode】Best Time to Buy and Sell 3 （hard) 自己做出来了 但别人的更好

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).

思路：

方案一：两次交易，得到最大收入。 设原本有length个数据，用n把数据分为[0 ~ n-1] 和 [n ~ length-1]两个部分，分别求最大收入，再加起来。结果超时了。

方案二：设数据是                        0  2  1  4  2  4  7

求后一个数字和前一个数字的差值：   2 -1  3  -2  2  3

把连续符合相同的数累加                 2 -1  3  -2   5

这样处理后，假设有m个数据， 用n把数据分为[0 ~ n-1] 和 [n ~ m-1]两个部分, 分别求两个部分的最大连续子段和。

由于经过预处理，数据变少了很多，所以就AC了。

int maxProfit3(vector<int> &prices) {
        if(prices.size() < )
        {
            return ;
        }

        //第一步：把prices中的数 两两间的差值算出来 把差值符号相同的加在一起
        vector<int> dealPrices;
        vector<int>::iterator it;
        int last = prices[];
        int current;
        int sumNum = ;
        for(it = prices.begin() + ; it < prices.end(); it++)
        {
            current = *it;
            if((current - last >=  && sumNum >= ) || (current - last <=  && sumNum <= ))
            {
                sumNum += current - last;
            }
            else
            {
                dealPrices.push_back(sumNum);
                sumNum = current - last;
            }
            last = current;
        }
        if(sumNum != )
        {
            dealPrices.push_back(sumNum);
        }

        //第二步
        if(dealPrices.size() == )
        {
            return dealPrices[] >  ? dealPrices[] : ;
        }
        else
        {
            int maxprofit = ;
            for(int n = ; n < dealPrices.size(); n++)
            {
                //求前半段最大连续子段和
                int maxSum1 = ;
                int maxSum2 = ;
                int curSum = ;
                for(int i = ; i < n; i++)
                {
                    curSum = (curSum > ) ? curSum + dealPrices[i] : dealPrices[i];
                    maxSum1 = (curSum > maxSum1) ? curSum : maxSum1;
                }

                //求后半段最大连续子段和
                curSum = ;
                for(int i = n; i < dealPrices.size(); i++)
                {
                    curSum = (curSum > ) ? curSum + dealPrices[i] : dealPrices[i];
                    maxSum2 = (curSum > maxSum2) ? curSum : maxSum2;
                }

                if(maxSum1 + maxSum2 > maxprofit)
                {
                    maxprofit = maxSum1 + maxSum2;
                }
            }
            return maxprofit;
        }

虽然我的AC了，但实际上还是个O(N^2)的算法，来看看大神们的O(N)代码。

第一种：https://oj.leetcode.com/discuss/14806/solution-sharing-commented-code-o-n-time-and-o-n-space

用两个数组left[],right[].

left记录当前值减去它前面的最小值的结果

right记录 当前值后面的最大值减去当前值的结果

把 left[i]+right[i+1] 针对所有的i遍历一遍 得到最大的值就是答案

public class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length < ) return ;//one of zero days, cannot sell
        // break the problem in to subproblems, what is the max profit if i decide to buy and sell one stock on or before day i
        // and the other stock after day i

        int[] left = new int[prices.length];//store the max profit so far for day [0,i] for i from 0 to n
        int[] right = new int[prices.length];//store the max profit so far for the days [i,n] for i from 0 to n
        int minl,maxprofit,maxr,profit;
        maxprofit = ;//lower bound on profit
        minl = Integer.MAX_VALUE;//minimum price so far for populating left array
        for(int i = ; i < left.length; i++){
            if (prices[i] < minl) minl = prices[i];//check if this price is the minimum price so far
            profit = prices[i] - minl;//get the profit of selling at current price having bought at min price so far
            if (profit > maxprofit) maxprofit = profit;//if the profit is greater than the profit so far, update the max profit
            left[i] = maxprofit;
        }
        maxprofit = ;//reset maxprofit to its lower bound
        maxr = Integer.MIN_VALUE;//maximum price so far for populating the right array
        //same line of reasoning as the above
        for(int i = left.length - ; i >= ; i--){
            if (prices[i] > maxr) maxr = prices[i];
            profit = maxr - prices[i];
            if (profit > maxprofit) maxprofit = profit;
            right[i] = maxprofit;
        }
        //get the best by combining the subproblems as described above
        int best = ;
        for(int i = ; i < prices.length - ; i++){
            if (left[i] + right[i+] > best) best = left[i] + right[i+];
        }
        best = best > maxprofit ? best : maxprofit;
        // in total 3 passes required and 2 extra arrays of size n
        return best;

    }
}

第二种：更厉害，泛化到了k次交易的情况 而且代码特别短

https://oj.leetcode.com/discuss/15153/a-clean-dp-solution-which-generalizes-to-k-transactions

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // 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
        if (prices.size() <= ) return ;
        else {
            int K = ; // number of max transation allowed
            int maxProf = ;
            vector<vector<int>> f(K+, vector<int>(prices.size(), ));
            for (int kk = ; kk <= K; kk++) {
                int tmpMax = f[kk-][] - prices[];
                for (int ii = ; ii < prices.size(); ii++) {
                    f[kk][ii] = max(f[kk][ii-], prices[ii] + tmpMax);
                    tmpMax = max(tmpMax, f[kk-][ii] - prices[ii]);
                    maxProf = max(f[kk][ii], maxProf);
                }
            }
            return maxProf;
        }
    }
};