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).
不是一个很typical的思路,因为只能有2次transaction,所以。。。从前扫一次,从后面扫一次,分别存在2个buffer里面,然后在扫一次buffer就好啦
class Solution { public: int maxProfit(vector<int> &prices) { // Start typing your C/C++ solution below // DO NOT write int main() function if(prices.empty()) return 0; int minVal = prices[0]; int result = 0; vector<int> inOrder; inOrder.push_back(0); vector<int> revOrder; revOrder.assign(prices.size(),0); for(int i = 1; i < prices.size(); i++){ if(prices[i] > minVal){ result = MAX(result, ( prices[i] - minVal)); }else{ minVal = prices[i]; } inOrder.push_back(result); } int maxVal = prices[prices.size()-1]; result = 0; for(int i = prices.size()-2; i >= 0; i--){ if(prices[i] < maxVal){ result = MAX(result, maxVal - prices[i] ); }else{ maxVal = prices[i]; } revOrder[i] = result; } result = 0; for (int i = 0; i < inOrder.size()-1; i++) { result = MAX(result, inOrder[i] + revOrder[i+1]); } result = MAX(result, revOrder[0]); return result; } };