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)的算法非常easy想到：
找寻一个点j，将原来的price[0..n-1]切割为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，非常多工作是反复的，在求price[0..j]的最大profit中已经做过了。
相似于Best Time to Buy and Sell Stock。能够在O(1)的时间从price[0..j]推出price[0..j+1]的最大profit。

可是怎样从price[j..n-1]推出price[j+1..n-1]？反过来思考，我们能够用O(1)的时间由price[j+1..n-1]推出price[j..n-1]。

终于算法：
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，相同已知r[i]，求r[i-1]也非常easy。
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。
* 參考：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;
     
    }
};