[leetcode] 188. Best Time to Buy and Sell Stock IV 解题报告

题目链接：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/

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 k transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

思路：当k >= len/2时，问题就退化成了可以交易任意次了，所以只要将任意两天股票差大于０的加起来即可．

当k < len/2时，可以记录k次交易每次买之后和卖以后最大的利润：

１．第ｉ次买操作买下当前股票之后剩下的最大利润为第(i-1)次卖掉股票之后的利润－当前的股票价格．状态转移方程为：

　　　　buy[i] = max(sell[i-1]- curPrice, buy[i]);

２．第ｉ次卖操作卖掉当前股票之后剩下的最大利润为第ｉ次买操作之后剩下的利润＋当前股票价格．状态转移方程为：

　　　　sell[i] = max(buy[i]+curPrice, sell[i]);

代码如下：

class Solution {public:    int maxProfit(int k, vector<int>& prices) {        if(prices.size() ==0) return 0;        int len = prices.size(), ans =0　;        if(k >= len/2)        {            for(int i = 1; i < len; i++)                 if(prices[i]-prices[i-1])>0) ans += prices[i]-prices[i-1];            return ans;         }        vector<int> buy(len+1, INT_MIN), sell(len+1, 0);        for(auto val: prices)        {            for(int i =1; i <= k; i++)            {                buy[i] = max(sell[i-1]-val, buy[i]);                sell[i] = max(buy[i]+val, sell[i]);            }        }        return sell[k];    }};

参考：https://leetcode.com/discuss/102303/clear-c-solution