**[Lintcode]Best Time to Buy and Sell Stock IV 买卖股票的最佳时机 IV Leetcode

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 ktransactions.

Example

Given prices = [4,4,6,1,1,4,2,5], and k = 2, return 6.

分析：使用二维数组，res[i][j]代表0～i中，交易j次的最大收益。

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])

因为在计算local[i][j]时，i-1,交易了j次，并且位置i因为和第j次交易相邻，可以并入第j次交易的情况，所以需要使用两个数组，一个数组local记录以i为最后一次卖出时，最大收益，和一个数组global，记录0到i范围内整体最大收益。这样local可以用来计算之前提到的交易合并问题。

以下方法运行内存超出规定大小。

class Solution {
    /**
     * @param k: An integer
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    public int maxProfit(int k, int[] prices) {
        if(prices == null || prices.length == 0) return 0;
        int[][] local = new int[prices.length][k + 1];
        int[][] global = new int[prices.length][k + 1];

        int res = 0;
        for(int j = 1; j <= k; j++) {
            for(int i = 1; i < prices.length; i++) {
                int diff = prices[i] - prices[i - 1];
            
                local[i][j] = Math.max(local[i - 1][j] + diff, 
                    global[i - 1][j - 1] + Math.max(0, diff));
                    
                global[i][j] = Math.max(global[i - 1][j], local[i][j]);
            }
        }
        return global[prices.length - 1][k];
    }
};

考虑使用两个滚动数组代替。滚动数组要确保覆盖后的值不会被用到。因此又引入了临时变量tmp。

现在内存符合要求，但是超时。继续改。

class Solution {
    /**
     * @param k: An integer
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    public int maxProfit(int k, int[] prices) {
        if(prices == null || prices.length == 0) return 0;
        int[] global = new int[prices.length];
        int[] local = new int[prices.length];
        
        int res = 0;
        for(int j = 1; j <= k; j++) {

            int tmp = global[0];
            for(int i = 1; i < prices.length; i++) {
                int diff = prices[i] - prices[i - 1];
                local[i] = Math.max(local[i - 1] + diff, 
                    tmp + Math.max(0, diff));
                
                tmp = global[i];
                global[i] = Math.max(global[i - 1], local[i]);
                if(global[i] > res) res = global[i];
            }
        }
        return res;
    }
};

当i小于j时，例如3个交易日需要进行4次交易的情况，可以以排除，所以给i加上限制。i>=j.  通过测试。

class Solution {
    /**
     * @param k: An integer
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    public int maxProfit(int k, int[] prices) {
        if(prices == null || prices.length == 0) return 0;
        int[] global = new int[prices.length];
        int[] local = new int[prices.length];
        //Memory Limit Exceeded，改用两个一维滚动数组
		
        int res = 0;
        for(int j = 1; j <= k; j++) {

            int tmp = global[0];
            for(int i = j; i < prices.length; i++) {
                int diff = prices[i] - prices[i - 1];
                local[i] = Math.max(local[i - 1] + diff, 
                    tmp + Math.max(0, diff));
                
                tmp = global[i];
                global[i] = Math.max(global[i - 1], local[i]);
                if(global[i] > res) res = global[i];
            }
        }
        return res;
    }
};