LeetCode解题报告（191）-- 188. Best Time to Buy and Sell Stock IV

Problem

You are given an integer array prices where prices[i] is the price of a given stock on the ith day.

Design an algorithm to find the maximum profit. You may complete at most k transactions.

Notice that you may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

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3

Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

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3

Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Constraints:

0 <= k <= 109
0 <= prices.length <= 1i4
0 <= prices[i] <= 1000

Analysis

这道题目是股票买卖系列题的第四题，题目的细节要求这里就不再叙述了，就是买入和卖出。这道题目不一样的地方在于，它对交易次数有限制，一买一卖算是一次交易。这里有两个限制条件，一个是天数，第二个是交易的次数。同类型的题目，不过加多了一个限制条件，dp由一维转为二维。定义两个dp状态，

local[i][j]：在到达第i天时最多可进行j次交易，并且最后一催交易在最后一天卖出的最大利润。
global[i][j]：在到达第i天时最多可进行j次交易的最大利润。

状态转移方程为：

local[i][j] = max(global[i - 1][j - 1] + max(diff, 0), local[i - 1][j] + diff)
global[i][j] = max(local[i][j], global[i - 1][j])

这里有一个坑需要注意，如果当最大交易次数k远大于prices的天数时，这道题目就退化成之前的题目了。因此就不需要用dp来求解了。

Solution

两个状态转移都是从上到下，从左到右的，所以用一个一维数组就可以了。在处理k远大于天数的情况，复用前面系列题的逻辑即可。

Code

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

Summary

有了前面几题的铺垫，再做这一题难度会降低不少。难点在于划分出局部最优和全局最优两个状态转移方程，同时还要考虑到一些特殊的case。这道题目的分享到这里，谢谢！