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];
    }
   
};