350.intersection-of-two-arrays-ii.cpp

/*
 * @lc app=leetcode id=350 lang=cpp
 *
 * [350] Intersection of Two Arrays II
 *
 * https://leetcode.com/problems/intersection-of-two-arrays-ii/description/
 *
 * algorithms
 * Easy (55.27%)
 * Likes:    4900
 * Dislikes: 743
 * Total Accepted:    859.8K
 * Total Submissions: 1.6M
 * Testcase Example:  '[1,2,2,1]\n[2,2]'
 *
 * 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:
 *
 *
 * 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]
 * Explanation: [9,4] is also accepted.
 *
 *
 *
 * Constraints:
 *
 *
 * 1 <= nums1.length, nums2.length <= 1000
 * 0 <= nums1[i], nums2[i] <= 1000
 *
 *
 *
 * 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?
 *
 *
 */

// @lc code=start
class Solution
{
public:
    vector<int> intersect(vector<int> &nums1, vector<int> &nums2)
    {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        vector<int> res;
        for (vector<int>::iterator itrA = nums1.begin(), itrB = nums2.begin(); itrA != nums1.end() && itrB != nums2.end();)
        {
            if (*itrA == *itrB)
            {
                res.push_back(*itrA);
                itrA++;
                itrB++;
            }
            else if (*itrA < *itrB)
                itrA++;
            else
                itrB++;
        }
        return res;
    }
};
// @lc code=end