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?

解法一：

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int
 
>& nums2) {
        unordered_map<int, int> m;
        vector<int> res;
        for (auto a : nums1) ++m[a];
        for (auto a : nums2) {
            if (m[a]-- > 0) res.push_back(a);
        }
        return res;
    }
};

再來看一種方法，這種方法先給兩個陣列排序，然後用兩個指標分別指向兩個陣列的起始位置，如果兩個指標指的數字相等，則存入結果中，兩個指標均自增1，如果第一個指標指的數字大，則第二個指標自增1，反之亦然，參見程式碼如下：

解法二：

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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