LeetCode link: 349. Intersection of Two Arrays, 349. 两个数组的交集

中文题解

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must be unique and you may return the result in any order.

Difficulty: Easy

Input: nums1 = [1,2,2,1], nums2 = [2,2]

Output: [2]

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]

Output: [9,4] or [4,9]

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 1000

中文题解

1. Convert one of the arrays to a set. The elements are unique in a set.
2. When traversing the other array, if the an element is found to already exist in the set, it means that the element belongs to the intersection, and the element should be added to the results.
3. The results is also of set type because duplicate removal is required.

• Time: O(n).
• Space: O(n).

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        var results = new HashSet<Integer>();
        var num1Set = new HashSet<Integer>();

        for (var num : nums1) {
            num1Set.add(num);
        }

        for (var num : nums2) {
            if (num1Set.contains(num)) {
                results.add(num);
            }
        }

        return results.stream().mapToInt(num -> num).toArray();
    }
}

class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        set_of_nums1 = set(nums1)
        results = set()

        for num in nums2:
            if num in set_of_nums1:
                results.add(num)

        return list(results)

class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        unordered_set<int> results;
        unordered_set<int> set_of_nums1(nums1.begin(), nums1.end());

        for (auto num : nums2) {
            if (set_of_nums1.contains(num)) {
                results.insert(num);
            }
        }

        return vector<int>(results.begin(), results.end());
    }
};

var intersection = function (nums1, nums2) {
  let results = new Set()
  let num1Set = new Set(nums1)

  for (const num of nums2) {
    if (num1Set.has(num)) {
      results.add(num)
    }
  }

  return Array.from(results)
};

public class Solution
{
    public int[] Intersection(int[] nums1, int[] nums2)
    {
        var results = new HashSet<int>();
        var num1Set = new HashSet<int>();

        foreach (int num in nums1)
            num1Set.Add(num);

        foreach (int num in nums2)
        {
            if (num1Set.Contains(num))
            {
                results.Add(num);
            }
        }

        return results.ToArray();
    }
}

func intersection(nums1 []int, nums2 []int) []int {
    results := map[int]bool{}
    num1Set := map[int]bool{}

    for _, num := range nums1 {
        num1Set[num] = true
    }

    for _, num := range nums2 {
        if _, ok := num1Set[num]; ok {
            results[num] = true
        }
    }

    return slices.Collect(maps.Keys(results))
}

def intersection(nums1, nums2)
  set_of_nums1 = Set.new(nums1)
  results = Set.new

  nums2.each do |num|
    if set_of_nums1.include?(num)
      results << num
    end
  end

  results.to_a
end

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// Welcome to create a PR to complete the code of this language, thanks!

// Welcome to create a PR to complete the code of this language, thanks!

// Welcome to create a PR to complete the code of this language, thanks!

给定两个数组 nums1 和 nums2 ，返回 它们的 交集 。输出结果中的每个元素一定是 唯一 的。我们可以 不考虑输出结果的顺序 。

难度: 容易

输入: nums1 = [1,2,2,1], nums2 = [2,2]

输出: [2]

输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4]

输出: [9,4] 或者 [4,9]

1. 把其中一个数组转为set，数据结构set的特点是元素不重复。
2. 遍历另一个数组时，如果发现当前元素已经存在于set中，则说明该元素属于交集，将该元素加入结果集中。
3. 结果集也采用set类型，因为需要去重。