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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

羞愧啊，根本不用那么复杂！直接统计每段上升序列的最大差就能够了

class Solution {public:    int maxProfit(vector
     
       &prices) {        if( prices.size() < 2)            return 0;        int max_profit = 0;        for( int i = 1; i < prices.size(); ++i){            if( prices[i] - prices[i-1] > 0)                max_profit += prices[i] - prices[i-1];        }        return max_profit;    }};
     

（之前的做法，想太多了）

和best time to buy and sell stock的差别是同意多次买入和卖出，dp[i]保存当前能获得的最低股价

那么max_profit就须要累加prices[i]-dp[i]，这就会出现反复加的情况，比方prices 1 2 4相应dp 1 1 1，会反复加，所以每次算dp[i]要修正，取最大的prices[j]-dp[j]，其余的prices[j]=dp[j]

class Solution {public:    int maxProfit(vector
     
       &prices) {        if( prices.size() < 2)            return 0;        vector< int> dp( prices.size(), INT_MAX);        dp[0] = prices[0];        for( int i = 1; i < prices.size(); ++i){            int mini = prices[i];            for( int j = i - 1; j >= 0; --j){                if(prices[i] - dp[j] > prices[j] - dp[j]){//保证prices[i]-dp[i]最大                    mini = dp[j];                    dp[j] = prices[j];                    if( prices[j] == dp[j])                        break;                }                else{                    break;                }            }            dp[i] = mini;        }        int max_profit = 0;        for( int i = 0; i < prices.size(); ++i){            max_profit = prices[i] - dp[i] > 0 ? max_profit + prices[i] - dp[i] : max_profit;        }        return max_profit;    }};