LeetCode - Best Time to Buy and Sell Stock III 题解

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：

想O(N)的算法，

F[i]表示，在第i天卖出一次最大利润(1次)

G[i]表示，在第i天买入，后面某天卖出最大利润(1次)

结合F和G的结果即可

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        vector<int> &A = prices;
        int n = A.size();
        if(n <= 1)return 0;
        vector<int> F(n, 0), G(n, 0), K(n, 0);
        F[0] = 0;
        for(int i = 1; i < n; ++i){
            F[i] = max(0, max(F[i - 1] - A[i - 1] + A[i], A[i] - A[i - 1]));
           // cout << i << ':' << F[i] << endl;
        }
        //cout << endl;
        G[n - 1] = 0;
        for(int i = n - 2; i >= 0; --i){
            G[i] = max(0, max(G[i + 1] + A[i + 1] - A[i], A[i + 1] - A[i]));
            //cout << i << ':' << G[i] << endl;
        }
        //cout << endl;
        K[n - 1] = G[n - 1];
        for(int i = n - 2; i >= 0; --i){
            K[i] = max(G[i], K[i + 1]);
           // cout << i << ':' << K[i] << endl;
        }
        int ans = max(K[0], F[n - 1]);
        for(int i = 0; i < n - 1; ++i){
            ans = max(ans, F[i] + K[i + 1]);
        }
        return ans;
    }
};