【LeetCode】 2215. Find the Difference of Two Arrays

Description

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.

Constraints:

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

給予兩個初始值為 0 的陣列 nums1 和 nums2，回傳一個大小為 2 的 list：

• answer[0] 是一個 list，包含全部都不同的整數且指出現在 nums1 沒出現在 nums2
• answer[1] 是一個 list，包含全部都不同的整數且指出現在 nums2 沒出現在 nums1

在 list 中的整數可以以任意順序回傳。

限制：

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

Example:

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] = [].

Solution

• 先用兩個 set(nums_s[2]) 分別記錄 nums1 與 nums2 出現過哪些數字
  • 因為大小可能不同，需要分開跑兩次 for loop
• 再來就可以利用 set 去看那些數字只出現在自己這邊
  • 因為不能重複，建議也使用 set 儲存，最後再轉回 vector
• 效率與空間都使用了不少，應該還有改善的空間

Code

class Solution {
public:
    vector<vector<int>> findDifference(vector<int>& nums1, vector<int>& nums2) {
        vector<vector<int>> ans;
        set<int> diff_s[2];
        set<int> nums_s[2];

        for(int i = 0; i < nums1.size(); i++)
        {
            nums_s[0].insert(nums1[i]);
        }
        for(int i = 0; i < nums2.size(); i++)
        {
            nums_s[1].insert(nums2[i]);
        }

        for(int i = 0; i < nums1.size(); i++)
        {
            if(!nums_s[1].count(nums1[i]))
                diff_s[0].insert(nums1[i]);
        }
        for(int i = 0; i < nums2.size(); i++)
        {
            if(!nums_s[0].count(nums2[i]))
                diff_s[1].insert(nums2[i]);
        }

        ans.push_back(vector<int>(diff_s[0].begin(), diff_s[0].end()));
        ans.push_back(vector<int>(diff_s[1].begin(), diff_s[1].end()));
        return ans;
    }
};

tags: LeetCode C++