LeetCode OJ - 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).

考虑数组[1,2,4,2,5,7,2,4,9,0]

分析：考虑了半天终于有思路了，类似于Best Time to Buy and Sell Stock I，将数字分为两个区间[0, i]和[i+1 ， n-1]，分别求出两个区域的最大收益并相加即可，用迭代求出最后的收益。

方法一：i 取值选在每次的波峰位置，接着在[i+1 ， n-1]中找出最大收益，时间复杂度为O(n^2)

方法二：对于每一个i求出 [0, i]最大收益front[i], [i+1, n-1]最大收益back[i], 求出max{font[i] + back[i]}，这里有一些无效计算但是时间复杂度为O(n)。

class Solution {public:    int maxProfit(vector<int> &prices) {        int len = prices.size();        if(len < 2) return 0;        int *front = new int[len];         int *back = new int[len];                memset(front, 0, len * sizeof(int));        memset(back, 0, len * sizeof(int));        int minprice = prices[0];        for(int i = 1; i < len; i++) {            front[i] = prices[i] - minprice;            minprice = min(minprice, prices[i]);        }                int maxprice = prices[len - 1];        for(int i = len - 2; i >= 0; i--) {            back[i] = max(back[i+1], maxprice - prices[i+1]);   //back[i+1]代表历史最大            maxprice = max(maxprice, prices[i+1]);        }        int ret = 0;        for(int i = 0; i < len; i++) {            ret = max(ret, front[i] + back[i]);        }        return ret;    }};