LeetCode 题解（82）: Best Time to Buy and Sell Stock IV

题目：

Say you have an array for which the i^th 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).

题解：

需要用两个表来记录：a) 在i天内（包括i）进行至多j次交易的最大获利，用global[i][j]表示，及 b) 第i天卖出股票，完成至多j次交易的最大获利，用local[i][j]表示。

那么递推关系式为：

local[0][j] = global[0][j] = 0, for j = 0,...,k

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

global[i][j] = max(globals[i-1][j], local[i][j])

其中：

global[i-1][j-1] + prices[i]-prices[i-1] 表示：前i-1天完成了j-1笔交易的最大获利，加上，第i-1天买入并在第i天卖入的收入

local[i-1][j] + prices[i] - prices[i-1] 表示：前i-1已经完成了j笔交易，并且最后一次卖出是在第i-1天发生呢，那么我们把卖出时间推迟一天（由i-1推迟到i），则仍是j笔交易，但是利润变化为prices[i] - prices[i-1]，所以我们还要加上这一部分。

global的递推比较好理解：要么在i-1天完成j笔交易，要么第i天完成第j笔交易。

c++版：

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

Java版：

public class Solution {
    public int maxProfit(int k, int[] prices) {
        if(k > prices.length) {
            int gain = 0;
            for(int i = 1; i < prices.length; i++)
                if(prices[i] > prices[i-1])
                    gain += prices[i] - prices[i-1];
            return gain;
        }
        
        int[] global = new int[k+1];
        int[] local = new int[k+1];
        
        for(int i = 1; i < prices.length; i++) {
            for(int j = 1; j <= k; j++) {
                local[j] = Math.max(global[j-1] + prices[i] - prices[i-1], local[j] + prices[i] - prices[i-1]);
            }
            for(int j = 1; j <= k; j++) {
                global[j] = Math.max(global[j], local[j]);
            }
        }
        return global[k];
    }
}

Python版：

class Solution:
    # @return an integer as the maximum profit 
    def maxProfit(self, k, prices):
        if k > len(prices):
            profit = 0
            for i in range(1, len(prices)):
                if prices[i] > prices[i-1]:
                    profit += prices[i] - prices[i-1]
            return profit
        
        Global = [0] * (k+1)
        Local = [0] * (k+1)
        for i in range(1, len(prices)):
            for j in range(1, k+1):
                Local[j] = max(Global[j-1] + prices[i]-prices[i-1], Local[j] + prices[i] - prices[i-1])
            for j in range(1, k+1):
                Global[j] = max(Global[j], Local[j])
        return Global[k]