[ES6 Series] A Detailed Explanation of the Set Type

Set

1. Definition

A Set is a constructor used to generate a Set data structure, which can accept an array (or any other iterable data structure) as an argument for initialization. It allows storing values of any type, whether primitive or object references.

2. Features

1. Unique Values: A Set automatically removes duplicates and will not contain duplicate values. (The following rules apply for identity checks)
1. +0 and -0 are considered identical when checking for uniqueness, so they do not repeat.
2. undefined and undefined are identical, so they do not repeat.
3. NaN and NaN are not strictly identical, but in a Set, NaN is considered equal to NaN, so only one NaN can exist.

2. Any Data Type: Sets can store both primitive types (like strings and numbers) and reference types (like objects and functions).

3. Ordered: The values in a Set are ordered based on the insertion sequence.

4. Suitable for Set Operations: Sets efficiently support operations like intersections, unions, and differences.

3. Properties and Methods

const set = new Set();

// Adding values
set.add(1);
set.add(5);
set.add(1); // Duplicate values are ignored
set.add('hello');

// Checking if values exist
console.log(set.has(5));  // Output: true
console.log(set.has(10)); // Output: false

// Getting the Set size
console.log(set.size);    // Output: 3

// Iterating over Set
set.forEach(value => {
  console.log(value);
});
// Output:
// 1
// 5
// hello

// Deleting a value
set.delete(5);
console.log(set.size);    // Output: 2

// Clearing the Set
set.clear();
console.log(set.size);    // Output: 0

Advantages (Compared to Arrays)

Every item in a Set must be unique.

• Checking for elements

• Deleting elements: Elements can be removed directly using their value.

• Storing NaN: NaN cannot be found using indexOf() or includes() in arrays, but Set can store this value.

• Removing duplicates

• The time complexity for Set operations is O(1), while for arrays, it is O(n).

Applications

1. Array.from This method can convert a Set to an array.

2. Array Deduplication

const arr = [1, 2, 3, 3, 4, 4, 5];
const uniqueArr = [...new Set(arr)];
console.log(uniqueArr); // [1, 2, 3, 4, 5]

1. Checking if a value exists in a Set

const visitedPages = new Set();

function visitPage(page) {
  if (visitedPages.has(page)) {
    console.log(`${page} already visited.`);
  } else {
    visitedPages.add(page);
    console.log(`${page} is visited for the first time.`);
  }
}

// 示例
visitPage('home'); // 输出: home is visited for the first time.
visitPage('home'); // 输出: home already visited.

1. Implementing Union, Intersection, and Difference, such as for permission management or tag filtering.

const setA = new Set([1, 2, 3]);
const setB = new Set([2, 3, 4]);

// 并集
const union = new Set([...setA, ...setB]);
console.log(union); // 输出: Set(4) {1, 2, 3, 4}

// 交集
const intersection = new Set([...setA].filter(x => setB.has(x)));
console.log(intersection); // 输出: Set(2) {2, 3}

// 差集
const difference = new Set([...setA].filter(x => !setB.has(x)));
console.log(difference); // 输出: Set(1) {1}

const userPermissions = new Set(['read', 'write']);
const requiredPermissions = new Set(['read', 'delete']);

// 交集：用户实际拥有的权限
const validPermissions = new Set([...userPermissions].filter(x => requiredPermissions.has(x)));
console.log(validPermissions); // Set { 'read' }

// 差集：缺失的权限
const missingPermissions = new Set([...requiredPermissions].filter(x => !userPermissions.has(x)));
console.log(missingPermissions); // Set { 'delete' }

1. Data Subscription and Deduplication

const subscribers = new Set();

function subscribe(listener) {
  subscribers.add(listener); // 确保唯一
}

function notify(data) {
  subscribers.forEach(listener => listener(data));
}

// 示例
subscribe(data => console.log(`Listener 1 received: ${data}`));
subscribe(data => console.log(`Listener 2 received: ${data}`));
notify('Event fired');
// 输出:
// Listener 1 received: Event fired
// Listener 2 received: Event fired

1. Form Data Validation

const emailSet = new Set();

function addEmail(email) {
  if (emailSet.has(email)) {
    console.log('This email is already registered.');
  } else {
    emailSet.add(email);
    console.log('Email registered successfully.');
  }
}

// 示例
addEmail('test@example.com'); // Email registered successfully.
addEmail('test@example.com'); // This email is already registered.

Leetcode Problems Using Se

var containsDuplicate = function(nums) {
  const set = new Set();
  for (const num of nums) {
    if (set.has(num)) {
      return true;
    }
    set.add(num);
  }
  return false;
};

var intersection = function(nums1, nums2) {
  const set1 = new Set(nums1);
  const set2 = new Set(nums2);
  return [...set1].filter(num => set2.has(num));
};

var longestConsecutive = function(nums) {
  const set = new Set(nums);
  let maxLength = 0;

  for (const num of set) {
    if (!set.has(num - 1)) { // 只有当 num 是序列的起点时才检查
      let currentNum = num;
      let currentLength = 1;

      while (set.has(currentNum + 1)) {
        currentNum += 1;
        currentLength += 1;
      }
      maxLength = Math.max(maxLength, currentLength);
    }
  }

  return maxLength;
};