LeetCode OJ 123 Best Time to Buy and Sell Stock III

题目：

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).

难度： hard

思路：先用II中贪心算法求解，讲贪心算法进行的交易数与k进行比较，如果小k<=贪心算法的交易数，直接返回贪心算法的结果，否则进行DP。递推公式参考

http://segmentfault.com/a/1190000002565570

代码如下

class Solution {public:    int maxProfit(int k, vector<int>& prices) {        if(prices.empty()||prices.size()<2)            return 0;        int maxProfit=0;        int i=0;        int time=0;        while(i<prices.size()-1){            if(prices[i+1]>prices[i]){                maxProfit=maxProfit+prices[i+1]-prices[i];                time++;            }            i++;                  }               if(time<=k)            return maxProfit;        int local[k+1][prices.size()];        int global[k+1][prices.size()];        for(int i=0;i<=k;i++)            for(int j=0;j<prices.size();j++){                local[i][j]=0;                global[i][j]=0;            }        for(int i=1;i<=k;i++){            for(int j=1;j<prices.size();j++){                int diff=prices[j]-prices[j-1];                local[i][j] = max(global[i-1][j-1],local[i][j-1]+diff);                global[i][j] = max(global[i][j-1],local[i][j]);                            }        }        return global[k][prices.size()-1];    }   };