Best Time to Buy and Sell Stock III

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

思路

以为可以借用 只买一次的那个代码，但是怎么也不能得到完全正确的答案，所以还是有问题
看了别人的代码

1. 用两个同样大小的数组，分别记录在当前之前最小的价格，和当前最大的价格
2. 然后思路就是，在当前位置交易，之前的最大值+之后的最大

代码

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if(n == 0) return 0;
        vector<int> min_p(n), max_p(n);

        min_p[0] = prices[0];
        for(int i = 1; i < n; i ++) min_p[i] = min(min_p[i-1], prices[i]);

        max_p[n-1] = prices[n-1];
        for(int i = n - 2; i >= 0; i --) max_p[i] = max(max_p[i+1], prices[i]);

        int ret = 0;
        int tmp = 0;
        for(int i = 0; i < n; i ++){
            tmp = max(tmp, prices[i] - min_p[i]);
            ret = max(ret, tmp + max_p[i] - prices[i]);
        }
        return ret; 
        }
};