350.Intersection of Two Arrays II

Tags: Easy Hash Table Two Pointer

Links: https://leetcode.com/problems/intersection-of-two-arrays-ii/

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

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]

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?

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class Solution { 
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        int n1 = nums1.size(), n2 = nums2.size();
        vector<int> result;
        if (n1 == 0 || n2 == 0) return result;

        map<int, int> m;
        if (n1 < n2) {
            for (auto e : nums2){
                ++m[e];
            }
            for (auto e : nums1) {
                if (m.find(e) != m.end() && m[e] > 0) {
                    result.push_back(e);
                    --m[e];
                }
            }
        }
        else {
            for (auto e : nums1){
                ++m[e];
            }
            for (auto e : nums2) {
                if (m.find(e) != m.end() && m[e] > 0) {
                    result.push_back(e);
                    --m[e];
                }
            }
        }

        return result;
    }
};

第一种方法，建立hash table，方法，很好理解，但是速度太慢。

第二种方法，双指针，先排序，从头比较。

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res;
        int i = 0, j = 0;
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        while (i < nums1.size() && j < nums2.size()) {
            if (nums1[i] == nums2[j]) {
                res.push_back(nums1[i]);
                ++i; ++j;
            } else if (nums1[i] > nums2[j]) {
                ++j;
            } else {
                ++i;
            }
        }
        return res;
    }
};

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