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
     
       &prices) {        if(prices.size() < 2)        {            return 0;        }        //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
      
        dealPrices;        vector
       
        ::iterator it;        int last = prices[0];        int current;        int sumNum = 0;        for(it = prices.begin() + 1; it < prices.end(); it++)        {            current = *it;            if((current - last >= 0 && sumNum >= 0) || (current - last <= 0 && sumNum <= 0))            {                sumNum += current - last;            }            else            {                dealPrices.push_back(sumNum);                sumNum = current - last;            }            last = current;        }        if(sumNum != 0)        {            dealPrices.push_back(sumNum);        }        //第二步        if(dealPrices.size() == 1)        {            return dealPrices[0] > 0 ? dealPrices[0] : 0;        }        else        {            int maxprofit = 0;            for(int n = 1; n < dealPrices.size(); n++)            {                //求前半段最大连续子段和                int maxSum1 = 0;                int maxSum2 = 0;                int curSum = 0;                for(int i = 0; i < n; i++)                {                    curSum = (curSum > 0) ? curSum + dealPrices[i] : dealPrices[i];                    maxSum1 = (curSum > maxSum1) ? curSum : maxSum1;                }                //求后半段最大连续子段和                curSum = 0;                for(int i = n; i < dealPrices.size(); i++)                {                    curSum = (curSum > 0) ? 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：

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 < 2) return 0;//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 = 0;//lower bound on profit        minl = Integer.MAX_VALUE;//minimum price so far for populating left array        for(int i = 0; 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 = 0;//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 - 1; i >= 0; 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 = 0;        for(int i = 0; i < prices.length - 1; i++){            if (left[i] + right[i+1] > best) best = left[i] + right[i+1];        }        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

class Solution {public:    int maxProfit(vector
     
       &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() <= 1) return 0;        else {            int K = 2; // number of max transation allowed            int maxProf = 0;            vector
      
       
        > f(K+1, vector
        
         (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;        }    }};