[LeetCode-123] Best Time to Buy and Sell Stock III

Best Time to Buy and Sell Stock III

Say you have an array for which the i^th 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).

 

储存0-N天的最大收益和N-END天的最大收益就好。

 

class Solution {
    public:
        int maxProfit(vector<int> &prices) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            int day_cnt = prices.size();
            if (day_cnt < 2) {
                return 0;
            }
            vector<vector<int> > profits;
            vector<int> profits_02N(day_cnt, 0);
            vector<int> profits_N2E(day_cnt, 0);
            int last_min = INT_MAX, last_max = 0, tmp_profit = 0, max_profit = 0;
 
            for (int i = 0; i < day_cnt; ++i) {
                if (prices.at(i) < last_min) {
                    last_min = prices.at(i);
                } else {
                    tmp_profit = max<int>(tmp_profit, prices.at(i) - last_min);
                }  
                profits_02N[i] = tmp_profit;
            }  
            tmp_profit = 0;
            for (int i = day_cnt - 1; i >= 0; --i) {
                if (prices.at(i) > last_max) {
                    last_max = prices.at(i);
                } else {
                    tmp_profit = max<int>(tmp_profit, last_max - prices.at(i));
                }
                profits_N2E[i] = tmp_profit;
            }
           
            max_profit = profits_02N[day_cnt - 1];
            for (int i = 0; i < day_cnt - 1; ++i) {
                tmp_profit = profits_02N[i] + profits_N2E[i + 1];;
                if (tmp_profit > max_profit) {
                    max_profit = tmp_profit;
                }
            }
            return max_profit;
        }
};