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‎README.adoc

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@@ -4961,13 +4961,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
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//|{counter:codes}
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//|{leetcode_base_url}/design-hashset/[705. Design HashSet^]
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//|{source_base_url}/_0705_DesignHashSet.java[Java]
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//|{doc_base_url}/0705-design-hashset.adoc[题解]
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//|Easy
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/design-hashset/[705. Design HashSet^]
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|{source_base_url}/_0705_DesignHashSet.java[Java]
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|{doc_base_url}/0705-design-hashset.adoc[题解]
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|Easy
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|
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//|{counter:codes}
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//|{leetcode_base_url}/design-hashmap/[706. Design HashMap^]
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//|{source_base_url}/_0706_DesignHashMap.java[Java]

‎docs/0705-design-hashset.adoc

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[#0705-design-hashset]
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= 705. Design HashSet
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= 705. 设计哈希集合
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{leetcode}/problems/design-hashset/[LeetCode - Design HashSet^]
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https://leetcode.cn/problems/design-hashset/[LeetCode - 705. 设计哈希集合 ^]
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Design a HashSet without using any built-in hash table libraries.
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不使用任何内建的哈希表库设计一个哈希集合（HashSet）。
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To be specific, your design should include these functions:
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实现 `MyHashSet` 类：
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* `void add(key)` 向哈希集合中插入值 `key` 。
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* `bool contains(key)` 返回哈希集合中是否存在这个值 `key` 。
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* `void remove(key)` 将给定值 `key`
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从哈希集合中删除。如果哈希集合中没有这个值，什么也不做。
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* `add(value)`: Insert a value into the HashSet.
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* `contains(value)` : Return whether the value exists in the HashSet or not.
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* `remove(value)`: Remove a value in the HashSet. If the value does not exist in the HashSet, do nothing.
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*示例：*
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....
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输入：
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["MyHashSet", "add", "add", "contains", "contains", "add", "contains", "remove", "contains"]
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[[], [1], [2], [1], [3], [2], [2], [2], [2]]
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输出：
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[null, null, null, true, false, null, true, null, false]
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解释：
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MyHashSet myHashSet = new MyHashSet();
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myHashSet.add(1); // set = [1]
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myHashSet.add(2); // set = [1, 2]
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myHashSet.contains(1); // 返回 True
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myHashSet.contains(3); // 返回 False ，（未找到）
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myHashSet.add(2); // set = [1, 2]
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myHashSet.contains(2); // 返回 True
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myHashSet.remove(2); // set = [1]
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myHashSet.contains(2); // 返回 False ，（已移除）
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....
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*提示：*
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*Example:*
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* `0 \<= key \<= 10^6^`
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* 最多调用 `10^4^` 次 `add`、`remove` 和 `contains`
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[subs="verbatim,quotes,macros"]
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----
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MyHashSet hashSet = new MyHashSet();
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hashSet.add(1);
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hashSet.add(2);
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hashSet.contains(1); // returns true
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hashSet.contains(3); // returns false (not found)
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hashSet.add(2);
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hashSet.contains(2); // returns true
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hashSet.remove(2);
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hashSet.contains(2); // returns false (already removed)
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----
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*Note:*
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== 思路分析
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* All values will be in the range of `[0, 1000000]`.
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* The number of operations will be in the range of `[1, 10000]`.
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* Please do not use the built-in HashSet library.
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三种解决冲突的办法：
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. 拉链法
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. 开放地址法
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. 再哈希法
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image::images/0705-01.png[{image_attr}]
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[[src-0705]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0705_DesignHashSet.java[tag=answer]
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include::{sourcedir}/_0705_DesignHashset.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0705_DesignHashset_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/design-hashset/solutions/652778/she-ji-ha-xi-ji-he-by-leetcode-solution-xp4t/[705. 设计哈希集合 - 官方题解^]
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. https://leetcode.cn/problems/design-hashset/solutions/653252/xiang-jie-hashset-de-she-ji-zai-shi-jian-4plc/[705. 设计哈希集合 - 详解 HashSet 的设计：在时间和空间上做权衡^]
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. https://leetcode.cn/problems/design-hashset/solutions/653184/yi-ti-san-jie-jian-dan-shu-zu-lian-biao-nj3dg/[705. 设计哈希集合 - 【宫水三叶】一题三解：「简单数组」&「链表」& 「分桶数组」^]
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‎docs/images/0705-01.png

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‎docs/index.adoc

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// include::0705-design-hashset.adoc[leveloffset=+1]
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include::0705-design-hashset.adoc[leveloffset=+1]
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‎logbook/202503.adoc

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|{doc_base_url}/3115-maximum-prime-difference.adoc[题解]
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|✅ 质数判断
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|{counter:codes2503}
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|{leetcode_base_url}/design-hashset/[705. 设计哈希集合^]
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|{doc_base_url}/0705-design-hashset.adoc[题解]
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|✅ 使用开放地址法解决哈希冲突。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0705_DesignHashset.java

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package com.diguage.algo.leetcode;
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import java.util.Arrays;
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public class _0705_DesignHashset {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-05-10 19:50:48
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*/
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class MyHashSet {
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private int[] data;
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private int mask;
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private int size;
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public MyHashSet() {
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data = new int[8];
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Arrays.fill(data, -1);
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mask = 8 - 1;
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size = 0;
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}
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public void add(int key) {
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if (size >= data.length * 3 / 4) {
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resize(data.length * 2);
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}
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boolean result = add(data, key, mask);
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if (result) {
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size++;
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}
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}
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public void remove(int key) {
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int index = key & mask;
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boolean ok = false;
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for (int i = index; i < data.length; i++) {
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if (data[i] == key) {
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data[i] = -1;
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ok = true;
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}
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}
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if (!ok) {
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for (int i = 0; i < index; i++) {
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if (data[i] == key) {
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data[i] = -1;
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ok = true;
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}
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}
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}
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if (ok) {
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size--;
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}
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// 需要考虑在初始化时，没有值的情况的缩容问题
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if (size <= data.length / 3 && data.length > 8) {
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resize(data.length / 2);
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}
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}
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public boolean contains(int key) {
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int index = key & mask;
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for (int i = index; i < data.length; i++) {
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if (data[i] == key) {
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return true;
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}
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}
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for (int i = 0; i < index; i++) {
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if (data[i] == key) {
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return true;
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}
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}
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return false;
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}
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private void resize(int length) {
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int[] newData = new int[length];
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Arrays.fill(newData, -1);
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int newMask = length - 1;
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for (int k : data) {
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if (k == -1) {
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continue;
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}
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add(newData, k, newMask);
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}
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this.data = newData;
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this.mask = newMask;
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}
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private boolean add(int[] data, int key, int newMask) {
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int index = key & newMask;
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if (data[index] == -1) {
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data[index] = key;
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return true;
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} else if (data[index] == key) {
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return false;
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} else {
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// 使用开放地址法解决哈希冲突
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for (int j = index + 1; j < data.length; j++) {
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if (data[j] == -1) {
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data[j] = key;
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return true;
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} else if (data[j] == key) {
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return false;
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}
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}
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for (int j = 0; j < index; j++) {
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if (data[j] == -1) {
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data[j] = key;
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return true;
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} else if (data[j] == key) {
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return false;
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}
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}
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}
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return false;
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}
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}
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// end::answer[]
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}

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