LeetCode: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).

分析：这一题约束最多只能买卖两次股票，并且手上最多也只能持有一支股票。因为不能连续买入两次股票，所以买卖两次肯定分布在前后两个不同的区间。设p(i) = 区间[0,1,2...i]的最大利润 + 区间[i,i+1,....n-1]的最大利润（式子中两个区间内分别只能有一次买卖，这就是第一道题的问题），那么本题的最大利润 = max{p[0],p[1],p[2],...,p[n-1]}。根据第一题的算法2，我们可以求区间[0,1,2...i]的最大利润；同理可以从后往前扫描数组求区间[i,i+1,....n-1]的最大利润，其递归式如下：

dp[i-1] = max{dp[i], maxprices - prices[i-1]}  ,maxprices是区间[i,i+1,...,n-1]内的最高价格。   

因此两趟扫描数组就可以解决这个问题，代码如下：

class Solution {public:    int maxProfit(vector<int> &prices) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        const int len = prices.size();        if(len <= 1)return 0;        int maxFromHead[len];        maxFromHead[0] = 0;        int minprice = prices[0], maxprofit = 0;        for(int i = 1; i < len; i++)        {            minprice = min(prices[i-1], minprice);            if(maxprofit < prices[i] - minprice)                maxprofit = prices[i] - minprice;            maxFromHead[i] = maxprofit;        }        int maxprice = prices[len - 1];        int res = maxFromHead[len-1];        maxprofit = 0;        for(int i = len-2; i >=0; i--)        {            maxprice = max(maxprice, prices[i+1]);            if(maxprofit < maxprice - prices[i])                maxprofit = maxprice - prices[i];            if(res < maxFromHead[i] + maxprofit)                res = maxFromHead[i] + maxprofit;        }        return res;    }};

转自：http://www.cnblogs.com/TenosDoIt/p/3436457.html