solve 188.best-time-to-buy-and-sell-stock-iv

Author: andavid

vscode/188.best-time-to-buy-and-sell-stock-iv.java

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+/*
+ * @lc app=leetcode id=188 lang=java
+ *
+ * [188] Best Time to Buy and Sell Stock IV
+ *
+ * https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/
+ *
+ * algorithms
+ * Hard (25.90%)
+ * Total Accepted:    80.3K
+ * Total Submissions: 309.9K
+ * Testcase Example:  '2\n[2,4,1]'
+ *
+ * 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).
+ *
+ * Example 1:
+ *
+ *
+ * Input: [2,4,1], k = 2
+ * Output: 2
+ * Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit
+ * = 4-2 = 2.
+ *
+ *
+ * Example 2:
+ *
+ *
+ * Input: [3,2,6,5,0,3], k = 2
+ * 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.
+ *
+ */
+class Solution {
+  public int maxProfit(int k, int[] prices) {
+    int profit = 0;
+    int n = prices.length;
+
+    // 当 k 大于天数的一半时，基本可以等同于交易任意次
+    if (k >= n/2) {
+      for (int i = 1; i < n; i++) {
+        if (prices[i] > prices[i-1]) {
+          profit += prices[i] - prices[i-1];
+        }
+      }
+      return profit;
+    }
+
+    int dp[][] = new int[k+1][n];
+    for (int i = 1; i <= k; i++) {
+      int temp = -prices[0];
+      for (int j = 1; j < n; j++) {
+        // 交易 i 次，第 j 天的最大收益
+        dp[i][j] = Math.max(dp[i][j-1], temp + prices[j]);
+        // 交易 i 次，第 j 天买入股票时的最大收益
+        temp = Math.max(temp, dp[i-1][j-1] - prices[j]);
+      }
+    }
+
+    return dp[k][n-1];
+  }
+}
+