[LeetCode 解题报告]123. Best Time to Buy and Sell Stock III

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 (i.e., you must sell the stock before you buy again).

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

考察：dp； 

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty())
            return 0;
      
        vector<int> sum(prices.size(), 0);
        int buy = INT_MAX, res = 0;
        for (int i = 0; i < prices.size(); i ++) {
            buy = min(buy, prices[i]);
            res = max(res, prices[i]-buy);
            sum[i] += res;
        }
        
        buy = INT_MIN, res = 0;
        for (int i = prices.size()-1; i >= 0; i --) {
            buy = max(buy, prices[i]);
            res = max(res, buy-prices[i]);
            sum[i] += res;
        }
        
        return *(max_element(sum.begin(), sum.end()));
    }
};

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty())
            return 0;
        
        int res1 = 0, res2 = 0;
        int buy1 = INT_MIN, buy2 = INT_MIN;
        for (int i = 0; i < prices.size(); i ++) {
            buy1 = max(buy1, -prices[i]); // 第一次买完
            res1 = max(res1, buy1+prices[i]); // 第一次卖完
            buy2 = max(buy2, res1-prices[i]); // 第二次买完
            res2 = max(res2, buy2+prices[i]); // 第二次卖完
        }
        return res2;
        
    }
};