readme.md

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)
Example:

Input: [1,2,3,0,2]
Output: 3 
Explanation: transactions = [buy, sell, cooldown, buy, sell]

Solution

需要想好每个状态是如何转化的.

package leetcode309_BestTimetoBuyandSellStockwithCooldown;

class Solution {
  public int maxProfit(int[] prices) {
    if (prices.length <= 1) return 0;

    // dp[m][n] m is the maxprofit from in range[0,...m].
    // n is the status that hold the stock or not.
    // 0 is not hold the stock. 1 is holding the stock.
    int[][] dp = new int[prices.length][2];

    dp[0][0] = 0;
    dp[0][1] = -prices[0];
    for (int i = 1; i < prices.length; i++) {
      dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
      if (i >= 2) {
        // dp[i - 1][1] 是昨天就买了, 今天不动
        // dp[i - 1][0] - prices[i] 是昨天没有, 今天买.但是这种情况中的有一种条件不对.
        // 一种是昨天卖了, 需要cooldown, 那不能买, 需要排除.
        // 一种是今天买的, 昨天是cooldown, 前天是拥有股票, 然后卖了.
        dp[i][1] = Math.max(dp[i - 1][1], dp[i - 2][0]  - prices[i]);
      } else {
        dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
      }
    }

    return Math.max(dp[prices.length - 1][0], dp[prices.length - 1][1]);
  }
}

// [2,1,4]
// [1,2,3,0,2]
// [6,1,6,4,3,0,2]