2215.Find the Difference of Two Arrays

tags: `Easy`,`Array`,`Hash Table`

題目描述

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

範例

Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4,6]
Output: [[1,3],[4,6]]
Explanation:
For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].

Example 2:

Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
Output: [[3],[]]
Explanation:
For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
Every integer in nums2 is present in nums1. Therefore, answer[1] = [].

Constraints:

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

解答

TypeScript













function findDifference(nums1: number[], nums2: number[]): number[][] {
  const set1 = new Set(nums1);
  const set2 = new Set(nums2);

  set1.forEach((item) => {
    if (set2.has(item)) {
      set2.delete(item);
      set1.delete(item);
    }
  });

  return [[...set1], [...set2]];
}

SheepWed, May 3, 2023

Javascript













function findDifference(nums1, nums2) {
  const set1 = new Set(nums1);
  const set2 = new Set(nums2);

  for (const item of set1) {
    if (set2.has(item)) {
      set1.delete(item);
      set2.delete(item);
    }
  }

  return [[...set1], [...set2]];
}

慚愧，跟我弟學習了，本來用set1去刪set2然後再用set2刪set1，不止要做兩遍還要多建一個set根本87
MarsgoatWed, May 3, 2023

Python



class Solution:
    def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
        return [list(set(nums1) - set(nums2)), list(set(nums2) - set(nums1))]

Ron ChenWed, May 3, 2023

C#













public IList<IList<int>> FindDifference(int[] nums1, int[] nums2) {
    HashSet<int> set1 = new HashSet<int>(nums1);
    HashSet<int> set2 = new HashSet<int>(nums2);

    foreach(int num in set2) {
        if (set1.Contains(num)) {
            set1.Remove(num);
            set2.Remove(num);
        }
    }
    
    return new List<IList<int>>(){ set1.ToList(), set2.ToList()};
}

JimWed, May 3, 2023

Java

























class Solution {
    public List<List<Integer>> findDifference(int[] nums1, int[] nums2) {
        Set<Integer> set1 = new HashSet<>();
        Set<Integer> set2 = new HashSet<>();

        for(Integer num : nums1) {
            set1.add(num);
        }

        for(Integer num : nums2) {
            set2.add(num);
        }

        Set<Integer> difference1 = new HashSet<>(set1);
        difference1.removeAll(set2);
        Set<Integer> difference2 = new HashSet<>(set2);
        difference2.removeAll(set1);

        List<List<Integer>> ans = new ArrayList<>();
        ans.add(difference1.stream().toList());
        ans.add(difference2.stream().toList());

        return ans;
    }
}

只能醜醜寫的我
Ron ChenWed, May 3, 2023













class Solution {
    public List<List<Integer>> findDifference(int[] nums1, int[] nums2) {
        Set<Integer> set1 = Arrays.stream(nums1).boxed().collect(Collectors.toSet());

        // Delete nums2 element which in set1 and assign to set2
        Set<Integer> set2 = Arrays.stream(nums2).filter(n -> !set1.contains(n)).boxed().collect(Collectors.toSet());

        // Delete set1 element which in nums2
        Arrays.stream(nums2).forEach(set1::remove);

        return Arrays.asList(new ArrayList<>(set1), new ArrayList<>(set2));
    }
}

解答區看到的 Java Stream API 用法，太猛了

Reference

回到題目列表