算法设计与应用基础

188. 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).

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

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假设你有一个数组，其中i元素是给定股票在第一天的价格。
设计利润最大化算法。您最多可以完成k个事务。
注：
你不能同时进行多个交易（你必须在卖出股票之前再购买）。

这题比较有难度，需要熟练掌握向量的运用才能解答

class Solution {


public:
    int maxProfit(int k, vector<int> &prices) {
        int len = prices.size();
        if (len<2) return 0;
        if (k>len/2){ // simple case
            int ans = 0;
            for (int i=1; i<len; ++i){
                ans += max(prices[i] - prices[i-1],0);
            }
            return ans;
        }
        int hold[k+1];
        int rele[k+1];
        for (int i=0;i<=k;++i){
            hold[i] = INT_MIN;
            rele[i] = 0;
        }
        int cur;
        for (int i=0; i<len; ++i){
            cur = prices[i];
            for(int j=k; j>0; --j){
                rele[j] = max(rele[j],hold[j] + cur);
                hold[j] = max(hold[j],rele[j-1] - cur);
            }
        }
        return rele[k];
    }
};