Intersection of Two Arrays II

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350.Intersection of Two Arrays II

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:

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

Example 2:

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Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.

双指针：

时间复杂度：O(mlogm+nlogn)

空间复杂度：O(1)

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class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        int length1 = nums1.size(), length2 = nums2.size();
        int index1 = 0, index2 = 0;
        vector<int> intersection;
        while (index1 < length1 && index2 < length2) {
            int num1 = nums1[index1], num2 = nums2[index2];
            if (num1 == num2) {
                intersection.push_back(num1);
                index1++;
                index2++;
            } else if (num1 < num2) {
                index1++;
            } else {
                index2++;
            }
        }
        return intersection;
    }
};

哈希表：

时间复杂度：O(m+n)

空间复杂度：O(min(m,n))

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class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        if (nums1.size() > nums2.size()) {
            return intersect(nums2, nums1);
        }
        unordered_map <int, int> m;
        for (int num : nums1) {
            ++m[num];
        }
        vector<int> intersection;
        for (int num : nums2) {
            if (m.count(num)) {
                intersection.push_back(num);
                --m[num];
                if (m[num] == 0) {
                    m.erase(num);
                }
            }
        }
        return intersection;
    }
};