[LeetCode] 188. Best Time to Buy and Sell Stock IV

 轉自LeetCode

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

Find the maximum profit you can achieve. You may complete at most k transactions.

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

 

Example 1:

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:

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 <= 100
• 0 <= prices.length <= 1000
• 0 <= prices[i] <= 1000

Solution

Best Time to Buy and Sell Stock 系列中的 hard 難題

也是得用 DP 解

dp[i][j] : max profit from at most i transaction using prices [0..j]

因為一定要遵守先買後賣的原則，所以一開始 currProfit 都會是 -prices[0]

其餘部分，可以看註解

kotlin(參考資料)

	class Solution {
	fun maxProfit(k: Int, prices: IntArray): Int {
	return if(prices.isEmpty()) {
	0
	} else {
	// dp[i][j]: max profit from at most i transactions using prices[0..j]
	val dp = Array(k+1) { IntArray(prices.size) {0}}
	var currProfit: Int // local max profit
	for(i in 1..k) {
	currProfit = -prices[0] // you need to buy before selling
	for(j in 1..prices.lastIndex) {
	// max(do nothing, sell today), sell won't increase transaction count
	dp[i][j] = Math.max(dp[i][j-1], currProfit + prices[j])
	// max(currProfit, buy today), this value is used for the next loop
	// we need to buy before selling, so we check here should we buy today
	// we need to use dp[i-1][j-1] because buy a stock increase the transaction count
	currProfit = Math.max(currProfit, dp[i-1][j-1] - prices[j])
	}
	}
	dp[k][prices.lastIndex]
	}
	}
	}

view raw best_time_to_buy_and_sell_stock_IV.kt hosted with ❤ by GitHub