Leetcode Best Time to Buy and Sell Stock III

/*** 题目：https://oj.leetcode.com/problems/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).O(n^2)的算法很容易想到：找寻一个点j，将原来的price[0..n-1]分割为price[0..j]和price[j..n-1]，分别求两段的最大profit。进行优化：对于点j+1，求price[0..j+1]的最大profit时，很多工作是重复的，在求price[0..j]的最大profit中已经做过了。类似于Best Time to Buy and Sell Stock，可以在O(1)的时间从price[0..j]推出price[0..j+1]的最大profit。但是如何从price[j..n-1]推出price[j+1..n-1]？反过来思考，我们可以用O(1)的时间由price[j+1..n-1]推出price[j..n-1]。最终算法：数组l[i]记录了price[0..i]的最大profit，数组r[i]记录了price[i..n]的最大profit。已知l[i]，求l[i+1]是简单的，同样已知r[i]，求r[i-1]也很容易。最后，我们再用O(n)的时间找出最大的l[i]+r[i]，即为题目所求。* 参考：http://blog.csdn.net/pickless/article/details/12034365* */class Solution {public:    int maxProfit(vector<int> &prices) {        const int size = prices.size();                if(size <= 1){            return 0;        }                int profitLeft[size];        int profitRight[size];                int profit = 0;        int i = 0;        int min = 0;        int max = 0;                memset(profitLeft, 0, sizeof(int) * size);        memset(profitRight, 0, sizeof(int) * size);                //store the maxProfit into profitLeft[i], from left to right        //profit = 0;        min = prices[0];        for(i = 1; i < size; i++){            //profitLeft[i] = (prices[i] - min) > profitLeft[i -1] ? (prices[i] - min) : profitLeft[i - 1];            //min = prices[i] < min ? prices[i] : min;            if(prices[i] - min > profitLeft[i -1]){//注意，此处不要写成if(prices[i] - min > profit){                profitLeft[i] = prices[i] - min;            } else {                profitLeft[i] = profitLeft[i - 1];            }                        if(prices[i] < min){                min = prices[i];            }        }                //store the maxProfit into profitRight[i], from right to left       // profit = 0;        max = prices[size - 1];        for(i = size - 2; i >= 0; i--){                       if(max - prices[i] > profitRight[i + 1]){//注意，此处不要写成if(max - prices[i] > profit){                profitRight[i] = max - prices[i];            } else {                profitRight[i] = profitRight[i + 1];            }                        if(prices[i] > max){                max = prices[i];            }        }                profit = profitRight[0] + profitLeft[0];        for(i = 1; i < size; i++){            if(profitRight[i] + profitLeft[i] > profit){                profit = profitRight[i] + profitLeft[i];             }         }                return profit;         }};