LeetCode 188 - Best Time to Buy and Sell Stock IV
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Say you have an array for which the ith element is the price of a given stock on dayi.
Design an algorithm to find the maximum profit. You may complete at mostk transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
解答: 本题在之前的基础上又加入了次数的限制。当次数大于数组长度的一般时,交易可以覆盖所有的数组元素,所以可以按照无限多次交易处理。其他情况,K小于数组的一半,还是最好使用动态规划,最重要的是在加入次数K限制的时候,如何建立状态转移函数。在之前题目的基础上,加入次数K.
- hold[i][j] :对于0-i天中最多交易 j 次并且第 i 天仍然持有股票的收益;
- unhold[i][j]:对于0到i天中最多交易 j次 并且第 i 天 不持有股票的最大收益。
- 动态规划状态转换如代码所示。
public class Solution { public int maxProfit(int k, int[] prices) { int n = prices.length; if(k == 0 || prices == null || n < 2){ return 0; } if(k >= n/2){ return nolimit(prices); } int[][] hold = new int[k+1][n]; int[][] unhold = new int[k+1][n]; for(int i = 1; i <= k; i++){ hold[i][0] = -prices[0]; unhold[i][0] = 0; for(int j = 1; j < n; j++){ hold[i][j] = Math.max(unhold[i-1][j] - prices[j],hold[i][j-1]); unhold[i][j] = Math.max(hold[i][j-1] + prices[j],unhold[i][j-1]); } } return unhold[k][n-1]; } public int nolimit(int[] prices){ int res = 0; for(int i = 1; i < prices.length; i++){ if(prices[i] > prices[i-1]) res += (prices[i] - prices[i-1]); } return res; } }
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