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() < 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<int> dealPrices; vector<int>::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：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 < 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
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() <= 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; } } };