Leetcode - Best Time to Buy and Sell Stock IV

题目链接

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).

解答思路

TODO （稍后补充）

解题代码

class Solution {
public:
    int getMaxProfileWhenKGtLen(vector& prices) {
        int maxProfit = 0;
        for (int i=1;i=0 ? (maxProfit+prices[i]-prices[i-1]) : maxProfit;
        }
        return maxProfit;
    }
    int maxProfit(int k, vector& prices) {
        
        if (prices.empty()) return 0;
        if (k >= prices.size()) return getMaxProfileWhenKGtLen(prices);
        
        /*
        // generate maxKProfit[prices.size()][k] and curKProfit[prices.size()][k]
        int **maxKProfit = new int*[prices.size()];
        int **curKProfit = new int*[prices.size()];
        for (int i=0;i > maxKProfit(prices.size(), vector(k+1, 0));
        vector > curKProfit(prices.size(), vector(k+1, 0));
        
        for (int i=1;i= maxKProfit[i-1][w-1]+diffI ? maxKProfit[i-1][w-1] : maxKProfit[i-1][w-1]+diffI;
                tmpMax = tmpMax >= curKProfit[i-1][w]+diffI ? tmpMax : curKProfit[i-1][w]+diffI;
                curKProfit[i][w] = tmpMax;
                maxKProfit[i][w] = maxKProfit[i-1][w] >= curKProfit[i][w] ? maxKProfit[i-1][w] : curKProfit[i][w];
            }
        }
        return maxKProfit[prices.size()-1][k];
        
    }
};

遇到Test Cases和Submission的结果不同的问题

如果用c++指针，会引起一些不确定性行为，所以尽量用vector吧，可以参考如下解释：

不确定性的解释