LeetCode 188. Best Time to Buy and Sell Stock IV

原题链接在这里：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/

题目：

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). 

题解：

若是k很大已经超过了prices.length的时候, 若是按照DP的方法做会浪费时间以及空间. 此时可以直接参照Best Time to Buy and Sell Stock II做法.

Let local[i][j] denotes local maximum profit of up when prices up to i, transaction up to j, the last transaction happend at i.

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

global[i-1][j-1] is global maximum profit up to prices i-1, transaction up to j-1. The last transaction happen at prices[i]. If could be diff or 0.

local[i-1][j] is local maximum profit up to pricesi-1, transaction up to j. It already perform j transaction, then the stock sold at prices[i-1] must be sold at prices[i].

Time Complexity: O(prices.length * k), k是最多交易次数.

Space: O(k).

AC Java:

public class Solution {
    public int maxProfit(int k, int[] prices) {
        if(prices == null || prices.length == 0 || k == 0){
            return 0;
        }
        if(k>=prices.length/2){
            int res = 0;
            for(int i = 1; i < prices.length; i++){
                res += Math.max(0, prices[i]-prices[i-1]);
            }
            return res;
        }
        
        int[] local = new int[k+1];
        int[] global = new int[k+1];
        for(int i = 1; i<prices.length; i++){
            int diff = prices[i] - prices[i-1];
            for(int j = k; j>=1; j--){
                local[j] = Math.max(global[j-1] + Math.max(diff,0), local[j] + diff);
                global[j] = Math.max(global[j], local[j]);
            }
        }
        return global[k];
    }
}

是Best Time to Buy and Sell Stock III的general情况.