Intersection of Two Arrays

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349.Intersection of Two Arrays

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must be unique 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]

Example 2:

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

哈希表：

时间复杂度：O(n+m)

空间复杂度：O(n+m)

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class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        vector<int> ans;
        unordered_set<int> set1;
        unordered_set<int> set2;
        for (int num : nums1)set1.insert(num);
        for (int num : nums2)set2.insert(num);
        if (set1.size() > set2.size())swap(set1, set2);
        for (int num : set1) {
            if (set2.count(num))ans.push_back(num);
        }
        return ans;
    }
};

双指针：

时间复杂度：O(nlogn+mlogm)

空间复杂度：O(nlogn+mlogm)

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class Solution {
public:
    vector<int> intersection(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) {
                // 保证加入元素的唯一性
                if (!intersection.size() || num1 != intersection.back()) {
                    intersection.push_back(num1);
                }
                index1++;
                index2++;
            } else if (num1 < num2) {
                index1++;
            } else {
                index2++;
            }
        }
        return intersection;
    }
};