一刷826

Author: diguage

README.adoc

@@ -5807,13 +5807,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 |Medium
 |
 
-//|{counter:codes}
-//|{leetcode_base_url}/most-profit-assigning-work/[826. Most Profit Assigning Work^]
-//|{source_base_url}/_0826_MostProfitAssigningWork.java[Java]
-//|{doc_base_url}/0826-most-profit-assigning-work.adoc[题解]
-//|Medium
-//|
-//
+|{counter:codes}
+|{leetcode_base_url}/most-profit-assigning-work/[826. Most Profit Assigning Work^]
+|{source_base_url}/_0826_MostProfitAssigningWork.java[Java]
+|{doc_base_url}/0826-most-profit-assigning-work.adoc[题解]
+|Medium
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/making-a-large-island/[827. Making A Large Island^]
 //|{source_base_url}/_0827_MakingALargeIsland.java[Java]

docs/0826-most-profit-assigning-work.adoc

@@ -1,40 +1,74 @@
 [#0826-most-profit-assigning-work]
-= 826. Most Profit Assigning Work
+= 826. 安排工作以达到最大收益
 
-{leetcode}/problems/most-profit-assigning-work/[LeetCode - Most Profit Assigning Work^]
+https://leetcode.cn/problems/most-profit-assigning-work/[LeetCode - 826. 安排工作以达到最大收益^]
 
-We have jobs: `difficulty[i]` is the difficulty of the `i`th job, and `profit[i]` is the profit of the `i`th job. 
+你有 `n` 个工作和 `m` 个工人。给定三个数组： `difficulty`, `profit` 和 `worker`，其中:
 
-Now we have some workers. `worker[i]` is the ability of the `i`th worker, which means that this worker can only complete a job with difficulty at most `worker[i]`. 
+* `difficulty[i]` 表示第 `i` 个工作的难度，`profit[i]` 表示第 `i` 个工作的收益。
+* `worker[i]` 是第 `i` 个工人的能力，即该工人只能完成难度小于等于 `worker[i]` 的工作。
 
-Every worker can be assigned at most one job, but one job can be completed multiple times.
+每个工人 *最多* 只能安排 *一个* 工作，但是一个工作可以 *完成多次* 。
 
-For example, if 3 people attempt the same job that pays $1, then the total profit will be $3.  If a worker cannot complete any job, his profit is $0.
+* 举个例子，如果 3 个工人都尝试完成一份报酬为 `$1` 的同样工作，那么总收益为 `$3` 。如果一个工人不能完成任何工作，他的收益为 `$0`。
 
-What is the most profit we can make?
+返回 _在把工人分配到工作岗位后，我们所能获得的最大利润_。
 
-*Example 1:*
+*示例 1：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
-*Output:* 100 
-*Explanation:* Workers are assigned jobs of difficulty [4,4,6,6] and they get profit of [20,20,30,30] seperately.
-----
+....
+输入: difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
+输出: 100
+解释: 工人被分配的工作难度是 [4,4,6,6] ，分别获得 [20,20,30,30] 的收益。
+....
+
+*示例 2:*
+
+....
+输入: difficulty = [85,47,57], profit = [24,66,99], worker = [40,25,25]
+输出: 0
+....
 
-*Notes:*
+*提示:*
 
+* `n == difficulty.length`
+* `n == profit.length`
+* `m == worker.length`
+* `1 \<= n, m \<= 10^4^`
+* `1 \<= difficulty[i], profit[i], worker[i] \<= 10^5^`
 
-* `1 <= difficulty.length = profit.length <= 10000`
-* `1 <= worker.length <= 10000`
-* `difficulty[i], profit[i], worker[i]`  are in range `[1, 10^5]`
 
 
+== 思路分析
 
+先将 `worker` 排序，然后对 `profit` 和 `difficulty` 排序： `profit` 优先，相同则再根据 `difficulty` 排序。从后向前遍历 `worker`，只要 `difficulty` 满足，优先取获利最大的工作。
+
+TIP: 有点蒙对的感觉。不如题解更有说服力。
 
 [[src-0826]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0826_MostProfitAssigningWork.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0826_MostProfitAssigningWork_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/most-profit-assigning-work/solutions/2780326/pai-xu-shuang-zhi-zhen-pythonjavacgojsru-gthg/[826. 安排工作以达到最大收益 - 排序+双指针^]
+. https://leetcode.cn/problems/most-profit-assigning-work/solutions/2776977/an-pai-gong-zuo-yi-da-dao-zui-da-shou-yi-c0s1/[826. 安排工作以达到最大收益 - 官方题解^]

docs/index.adoc

@@ -1744,7 +1744,7 @@ include::0824-goat-latin.adoc[leveloffset=+1]
 
 include::0825-friends-of-appropriate-ages.adoc[leveloffset=+1]
 
-// include::0826-most-profit-assigning-work.adoc[leveloffset=+1]
+include::0826-most-profit-assigning-work.adoc[leveloffset=+1]
 
 // include::0827-making-a-large-island.adoc[leveloffset=+1]
 

logbook/202601.adoc

@@ -1041,6 +1041,10 @@ endif::[]
 |{doc_base_url}/0825-friends-of-appropriate-ages.adoc[题解]
 |❌ 滑动窗口。要仔细分析题目要求，根据要求筛选出合适的解题策略。
 
+|{counter:codes}
+|{leetcode_base_url}/most-profit-assigning-work/[826. 安排工作以达到最大收益^]
+|{doc_base_url}/0826-most-profit-assigning-work.adoc[题解]
+|✅ 排序+双指针。对入参排序，优先获取获利最大的工作，累加即可。有点蒙对的感觉。不如题解更有说服力。
 
 
 

src/main/java/com/diguage/algo/leetcode/_0826_MostProfitAssigningWork.java

@@ -0,0 +1,44 @@
+package com.diguage.algo.leetcode;
+
+import java.util.Arrays;
+import java.util.Comparator;
+
+public class _0826_MostProfitAssigningWork {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2026-07-19 21:59:47
+   */
+  public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
+    Arrays.sort(worker);
+    int n = difficulty.length;
+    int m = worker.length;
+    Integer[] index = new Integer[n];
+    for (int i = 0; i < n; i++) {
+      index[i] = i;
+    }
+    // 这样能通过验证，但是感觉有点蒙对了的感觉。
+    Arrays.sort(index,
+      Comparator.comparingInt((Integer a) -> profit[a])
+        .thenComparingInt(a -> difficulty[a]));
+    int result = 0;
+    int idx = n - 1;
+    for (int i = m - 1; i >= 0; i--) {
+      while (idx >= 0 && difficulty[index[idx]] > worker[i]) {
+        idx--;
+      }
+      if (idx >= 0) {
+        result += profit[index[idx]];
+      }
+    }
+    return result;
+  }
+  // end::answer[]
+
+  static void main() {
+    new _0826_MostProfitAssigningWork()
+//      .maxProfitAssignment(new int[]{85, 47, 57}, new int[]{24, 66, 99}, new int[]{40, 25, 25});
+      .maxProfitAssignment(new int[]{68, 35, 52, 47, 86}, new int[]{67, 17, 1, 81, 3}, new int[]{92, 10, 85, 84, 82});
+  }
+}