[Leetcode] 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).

设置一个指针，分别求出前面的最大收益与后面的最大收益，将两者相加，可以提前计算好存在两个数组里以避免重复计算。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int n = prices.size();
        if (n == 0) return 0;
        vector<int> first(n, 0);
        vector<int> second(n, 0);
        int max_profit = 0, cur_min = prices[0], cur_max = prices[n-1];
        for (int i = 1; i < n; ++i) {
            max_profit = max(max_profit, prices[i] - cur_min);
            cur_min = min(cur_min, prices[i]);
            first[i] = max_profit;
        }
        max_profit = 0;
        for (int i = n - 2; i >= 0; --i) {
            max_profit = max(max_profit, cur_max - prices[i]);
            cur_max = max(cur_max, prices[i]);
            second[i] = max_profit;
        }
        max_profit = 0;
        for (int i = 0; i < n; ++i) {
            max_profit = max(max_profit, first[i] + second[i]);
        }
        return max_profit;
    }
};

 下面是Best Time to Buy and Sell Stock I的代码，有人说将问题转化为最大连续子序列和的问题，其实不用那样，只要保存当前最小值就可以解决问题。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int n = prices.size();
        if (n == 0) return 0;
        int max_profit = 0, cur_min = prices[0];
        for (int i = 0; i < n; ++i) {
            max_profit = max(max_profit, prices[i] - cur_min);
            cur_min = min(cur_min, prices[i]);
        }
        return max_profit;
    }
};