【刷题-LeetCode】123 Best Time to Buy and Sell Stock III

1. 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 为分界，左边是第一次交易能够获得的最大利润，右边是第二次，最后两边加起来取最大

Note：不能每次都计算一次，用数组存储能够获得的最大利润，否则会超时

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 每次取最值时针对全局的利润，设置4个变量：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); //以低价买入
            s1=max(s1,b1+x); //以高价卖出
            b2=max(b2,s1-x); //低价买入，即结余要最大
            s2=max(s2,b2+x); //高价卖出
        }
        return s2;
    }
};