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.

用local[i][j]表示到达第i天时，最多进行j次交易的局部最优解；用global[i][j]表示到达第i天时，最多进行j次的全局最优解。它们二者的关系如下（其中diff = prices[i] – prices[i – 1]）：

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

local[i][j]和global[i][j]的区别是：local[i][j]意味着在第i天一定有交易（卖出）发生，当第i天的价格高于第i-1天（即diff > 0）时，那么可以把这次交易（第i-1天买入第i天卖出）跟第i-1天的交易（卖出）合并为一次交易，即local[i][j]=local[i-1][j]+diff；当第i天的价格不高于第i-1天（即diff<=0）时，那么local[i][j]=global[i-1][j-1]+diff，而由于diff<=0，所以可写成local[i][j]=global[i-1][j-1]。global[i][j]就是我们所求的前i天最多进行k次交易的最大收益，可分为两种情况：如果第i天没有交易（卖出），那么global[i][j]=global[i-1][j]；如果第i天有交易（卖出），那么global[i][j]=local[i][j]。

//if(k >= days)  go to II
//k < days
//local[0][0] = 0; global[0][0]
//local[i][j] = Math.max(local[i-1][j] + differ, global[i-1][j-1]); 在第i天一定进行交易 diff >0
//global[i][j] = Math.max(local[i][j], global[i-1][j]); 在第i 天进行交易，或者不进行交易

public class Solution {
    public int maxProfit(int k, int[] prices) {
        if(prices == null || prices.length <= 1 || k==0) return 0;
        int days = prices.length;
        if(k >= days) return getMaxProfit(prices); // question II
        int local[][] = new int[days][k+1];
        int global[][] = new int[days][k+1];
        local[0][0] = 0;
        global[0][0] = 0;
        for(int i = 1; i < days ; i++){
            int diff = prices[i] - prices[i-1];
            for(int j = 1 ; j <= k ; j ++){
                local[i][j] = Math.max(local[i-1][j] + diff, global[i-1][j-1]);
                global[i][j] = Math.max(local[i][j], global[i-1][j]);
            }
        }
        return global[days-1][k];
    }
    public int getMaxProfit(int[] prices) {
        int total = 0;
        for(int i = 1; i < prices.length; i++){
            if(prices[i] > prices[i-1]) total += prices[i]-prices[i-1];
        }
        return total;
    }
}