三刷122

Author: diguage

docs/0122-best-time-to-buy-and-sell-stock-ii.adoc

@@ -1,37 +1,49 @@
+
 [#0122-best-time-to-buy-and-sell-stock-ii]
-= 122. Best Time to Buy and Sell Stock II
+= 122. 买卖股票的最佳时机 II
 
-{leetcode}/problems/best-time-to-buy-and-sell-stock-ii/[LeetCode - Best Time to Buy and Sell Stock II^]
+https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/[LeetCode - 122. 买卖股票的最佳时机 II ^]
 
-Say you have an array for which the _i^th^_ element is the price of a given stock on day _i_.
+给你一个整数数组 `prices` ，其中 `prices[i]` 表示某支股票第 `i` 天的价格。
 
-Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
+在每一天，你可以决定是否购买和/或出售股票。你在任何时候 *最多* 只能持有 *一股* 股票。你也可以先购买，然后在 *同一天* 出售。
 
-*Note:* You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
+返回 _你能获得的 *最大* 利润_ 。
 
-.Example 1:
-----
-Input: [7,1,5,3,6,4]
-Output: 7
-Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
-             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
-----
+*示例 1：*
+
+....
+输入：prices = [7,1,5,3,6,4]
+输出：7
+解释：在第 2 天（股票价格 = 1）的时候买入，在第 3 天（股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5 - 1 = 4。
+随后，在第 4 天（股票价格 = 3）的时候买入，在第 5 天（股票价格 = 6）的时候卖出, 这笔交易所能获得利润 = 6 - 3 = 3。
+最大总利润为 4 + 3 = 7 。
+....
+
+*示例 2：*
+
+....
+输入：prices = [1,2,3,4,5]
+输出：4
+解释：在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5 - 1 = 4。
+最大总利润为 4 。
+....
+
+*示例 3：*
+
+....
+输入：prices = [7,6,4,3,1]
+输出：0
+解释：在这种情况下, 交易无法获得正利润，所以不参与交易可以获得最大利润，最大利润为 0。
+....
+
+
+*提示：*
+
+* `1 \<= prices.length \<= 3 * 10^4^`
+* `0 \<= prices[i] \<= 10^4^`
 
-.Example 2:
-----
-Input: [1,2,3,4,5]
-Output: 4
-Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
-             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
-             engaging multiple transactions at the same time. You must sell before buying again.
-----
 
-.Example 3:
-----
-Input: [7,6,4,3,1]
-Output: 0
-Explanation: In this case, no transaction is done, i.e. max profit = 0.
-----
 
 == 思路分析
 
@@ -74,8 +86,18 @@ include::{sourcedir}/_0122_BestTimeToBuyAndSellStockII.java[tag=answer]
 include::{sourcedir}/_0122_BestTimeToBuyAndSellStockII_2.java[tag=answer]
 ----
 --
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0122_BestTimeToBuyAndSellStockIi_3.java[tag=answer]
+----
+--
 ====
 
+
 == 参考资料
 
 . https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-with-cooldown/solutions/8610/yi-ge-fang-fa-tuan-mie-6-dao-gu-piao-wen-ti-by-lab/[309. 买卖股票的最佳时机含冷冻期 - 一个方法团灭 6 道股票问题^]

logbook/202503.adoc

@@ -518,6 +518,11 @@ endif::[]
 |{doc_base_url}/0069-sqrtx.adoc[题解]
 |✅ 二分查找。使用一个单独变量来保存最后一个小于等于的中间值，那么即可直接获取答案。
 
+|{counter:codes2503}
+|{leetcode_base_url}/best-time-to-buy-and-sell-stock-ii/[122. 买卖股票的最佳时机 II^]
+|{doc_base_url}/0122-best-time-to-buy-and-sell-stock-ii.adoc[题解]
+|✅ 有差价就买卖，见好就收。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_0122_BestTimeToBuyAndSellStockIi_3.java

@@ -0,0 +1,20 @@
+package com.diguage.algo.leetcode;
+
+public class _0122_BestTimeToBuyAndSellStockIi_3 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-27 07:59:27
+   */
+  public int maxProfit(int[] prices) {
+    int result = 0;
+    for (int i = 1; i < prices.length; i++) {
+      if (prices[i - 1] < prices[i]) {
+        // 见好就收
+        result += prices[i] - prices[i - 1];
+      }
+    }
+    return result;
+  }
+  // end::answer[]
+}