三刷90

Author: diguage

Diff for: docs/0000-00-note.adoc

@@ -150,7 +150,7 @@ image::images/quick-sort-01.gif[{image_attr}]
 .. Sliding Window Median (hard)
 .. Maximize Capital (hard)
 .. Problem Challenge 1: Next Interval (hard)
-. xref:0000-24-subsets.adoc[Subsets 子集]
+. xref:0000-25-subsets.adoc[Subsets 子集]
 .. Subsets (easy)
 .. Subsets With Duplicates (easy)
 .. Permutations (medium)

Diff for: docs/0000-24-subsets.adoc renamed to docs/0000-25-subsets.adoc

@@ -1,4 +1,4 @@
-[#0000-24-subsets]
+[#0000-25-subsets]
 = Subsets 子集
 
 一般都是使用多重DFS
@@ -29,10 +29,3 @@
 . xref:0241-different-ways-to-add-parentheses.adoc[241. Different Ways to Add Parentheses]
 . xref:0320-generalized-abbreviation.adoc[320. Generalized Abbreviation]
 . xref:0784-letter-case-permutation.adoc[784. Letter Case Permutation]
-
-
-String Permutations by changing case (medium)
-
-Balanced Parentheses (hard)
-
-Unique Generalized Abbreviations (hard)

Diff for: docs/0090-subsets-ii.adoc

@@ -1,25 +1,30 @@
 [#0090-subsets-ii]
-= 90. Subsets II
+= 90. 子集 II
 
-{leetcode}/problems/subsets-ii/[LeetCode - Subsets II^]
+https://leetcode.cn/problems/subsets-ii/[LeetCode - 90. 子集 II ^]
 
-Given a collection of integers that might contain duplicates, `nums`, return all possible subsets (the power set).
+给你一个整数数组 `nums`，其中可能包含重复元素，请你返回该数组所有可能的 子集（幂集）。
 
-*Note:* The solution set must not contain duplicate subsets.
+解集 *不能* 包含重复的子集。返回的解集中，子集可以按 *任意顺序* 排列。
 
-.Example:
-----
-Input: [1,2,2]
-Output:
-[
-  [2],
-  [1],
-  [1,2,2],
-  [2,2],
-  [1,2],
-  []
-]
-----
+*示例 1：*
+
+....
+输入：nums = [1,2,2]
+输出：[[],[1],[1,2],[1,2,2],[2],[2,2]]
+....
+
+*示例 2：*
+
+....
+输入：nums = [0]
+输出：[[],[0]]
+....
+
+*提示：*
+
+* `+1 <= nums.length <= 10+`
+* `+-10 <= nums[i] <= 10+`
 
 == 解题分析
 
@@ -56,6 +61,15 @@ include::{sourcedir}/_0090_SubsetsII.java[tag=answer]
 include::{sourcedir}/_0090_SubsetsII_2.java[tag=answer]
 ----
 --
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0090_SubsetsII_3.java[tag=answer]
+----
+--
 ====
 
 == 参考资料

Diff for: docs/index.adoc

@@ -70,7 +70,7 @@ include::0000-23-transform-and-conquer.adoc[leveloffset=+1]
 
 include::0000-24-backtrack.adoc[leveloffset=+1]
 
-include::0000-24-subsets.adoc[leveloffset=+1]
+include::0000-25-subsets.adoc[leveloffset=+1]
 
 include::0000-25-greedy.adoc[leveloffset=+1]
 

Diff for: logbook/202503.adoc

@@ -205,6 +205,11 @@ endif::[]
 |{doc_base_url}/0322-coin-change.adoc[题解]
 |⭕️ 动态规划，完全背包问题
 
+|{counter:codes2503}
+|{leetcode_base_url}/subsets-ii/[90. 子集 II]
+|{doc_base_url}/0090-subsets-ii.adoc[题解]
+|⭕️ 子集，需要注意重复元素的处理。现在用 `Set` 记录已添加子集的方案还可以再优化。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

Diff for: src/main/java/com/diguage/algo/leetcode/_0090_SubsetsII_3.java

@@ -0,0 +1,38 @@
+package com.diguage.algo.leetcode;
+
+import java.util.*;
+
+public class _0090_SubsetsII_3 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-09 18:04:54
+   */
+  public List<List<Integer>> subsetsWithDup(int[] nums) {
+    List<List<Integer>> result = new ArrayList<>();
+    Arrays.sort(nums);
+    result.add(new ArrayList<>());
+    // 使用 Set 记录已添加的子集，这个方案还需要再优化
+    Set<String> existed = new HashSet<>();
+    for (int i = 0; i < nums.length; i++) {
+      int num = nums[i];
+      int size = result.size();
+      for (int j = 0; j < size; j++) {
+        List<Integer> subset = new ArrayList<>(result.get(j));
+        subset.add(num);
+        StringBuilder sb = new StringBuilder(nums.length);
+        for (Integer ai : subset) {
+          sb.append(ai);
+        }
+        if (existed.add(sb.toString())) {
+          result.add(subset);
+        }
+      }
+    }
+    return result;
+  }
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0090_SubsetsII_3().subsetsWithDup(new int[]{1, 2, 2});
+  }
+}