LeetCode刷题笔记：两个数组的交集 II

Given two arrays, write a function to compute their intersection.

Example:

Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

• Each element in the result should appear as many times as it shows in both arrays.
• The result can be in any order.

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?

解题思路

使用hash，首先将数组1的元素插入到一个map中，然后对数组2中的每个元素进行查找，时间复杂度为O（n），空间复杂度为O（n）

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {

        vector<int> result;


        int size1 = nums1.size();
        int size2 = nums2.size();

        map<int, int> map;


        for(int i = 0; i < size1; ++i) {
            map[nums1[i]] += 1;
        }

        for(int i = 0; i < size2; ++i) {
            if(map[nums2[i]] > 0) {
                result.push_back(nums2[i]);
                map[nums2[i]] -= 1;
            }
        }

        return result;
    }
};