Best Time to Buy and Sell Stock III - LeetCode 123

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).

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 Array Dynamic Programming

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分析：
最多买卖两次，那就把元素组分成两段，分别求出两段的最大利益，然后相加，于是得到动态规划解法（类似于求出数组中除去自身元素外的其他所有元素的乘积）：
先从前往后，求出第0天到第i天买卖一次能获得的最大利益，记入forward[i];
再从后往前，求出第i天到n-1 天买卖一次能获得的最大利益，记入backward[i];
然后两次买卖的最大利益 max_profi = max{forward[i] + backward[i]},其中i取值[0,n-1]

////////////////12ms//////////////////////////////////class Solution {public:    /**     * @param prices: Given an integer array     * @return: Maximum profit     */    int maxProfit(vector<int> &prices) {        // write your code here        int n = prices.size();        if(n < 2)            return 0;        if( n == 2)            return prices[1]> prices[0]? prices[1] - prices[0]:0;        vector<int> forward(n,0); //forward[i]表示从第0天到第i天买卖一次能获得的最大利益        vector<int> backward(n,0); //backward[i]表示从第i天到n-1天买卖一次能获得的最大利益                 int min_p = prices[0];        for(int i = 1; i < n; i++){  //从前往后计算forward             if(prices[i] < min_p)                min_p = prices[i];            int t = prices[i] - min_p;            forward[i] =  max(t,forward[i-1]);        }                int max_p = prices[n-1];        for(int i = n-2; i >= 0; i--){  //从后往前计算backward            if(prices[i] > max_p)                max_p = prices[i];            int t =  max_p - prices[i];            backward[i] = max(t,backward[i+1]);        }                int max_profi = 0;          for(int i = 0; i < n - 1; i++){ //求出两段和最大值            int t = forward[i] + backward[i];            max_profi = max(max_profi , t);        }                return max_profi;    }};