LeetCode 350. Intersection of Two Arrays II 解題紀錄

2022-01-24 299 words One minute

Contents

題目

Problem

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

Example 1:

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

Example 2:

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

Explanation: [9,4] is also accepted.

Constraints:

1 $\leq$ nums1.length, nums2.length $\leq$ 1000
0 $\leq$ nums1[i], nums2[i] $\leq$ 1000

Follow up:

What if the given array is already sorted? How would you optimize your algorithm?
What if nums1’s size is small compared to nums2’s size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

想法

確認兩個 array 中是否有 size 為 0 的情況，若有，則直接 return 空。
遍歷 array 中所有個數，紀錄到 unordered_map 中
另一個 array 則檢查是否對應的 key 值不為 0，表示前一個 array 中也有這個元素，故印出它。

解法

vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {

    // 檢查是否有 array 為空，若為空，則不可能有交集
    if(!nums1.size() || !nums2.size())  return {};

    vector<int> ans;

    // 紀錄每個數字出現的次數
    unordered_map<int, int> mp;

    for(auto i:nums1)
    {
        mp[i]++;
    }

    for(auto i:nums2)
    {
        // 若 mp[i] != 0 表示有交集
        if(mp[i])
        {
            ans.push_back(i);
            mp[i]--;
        }
    }

    return ans;
}

Time complexity: $\mathcal{O}(n)$.
Space complexity: $\mathcal{O}(n)$.