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.

Solution

Approach1 以 i 為分界，左邊是第一次交易能夠獲得的最大利潤，右邊是第二次，最後兩邊加起來取最大

Note：不能每次都計算一次，用陣列儲存能夠獲得的最大利潤，否則會超時

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int ans = 0;
        int n = prices.size();
        if(n == 0)return ans;
        int left[n] = {0}, right[n] = {0};
        int min_price = prices[0];
        for(int i = 1; i < n; ++i){
            min_price = min(min_price, prices[i]);
            left[i] = max(left[i-1], prices[i] - min_price);
        }
        int max_price = prices[n-1];
        for(int i = n-2; i >= 0; --i){
            max_price = max(max_price, prices[i]);
            right[i] = max(right[i+1], max_price - prices[i]);
        }
        for(int i = 0; i < n; ++i){
            ans = max(ans, left[i] + right[i]);
        }
        return ans;
    }
};
 

Appraoch 2 每次取最值時針對全域性的利潤，設定4個變數：b1, s1, b2, s2

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int b1 = INT_MIN, b2 = INT_MIN;
        int s1 = 0, s2 = 0;
        for(int x : prices){
            b1=max(b1,-x); //以低價買入
            s1=max(s1,b1+x); //以高價賣出
            b2=max(b2,s1-x); //低價買入，即結餘要最大
            s2=max(s2,b2+x); //高價賣出
        }
        return s2;
    }
};