350. Intersection of Two Arrays II

Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.

Solution1：Hashmap

思路： Keep available 数量
Time Complexity: O(N) Space Complexity: O(N)
也可以是extraO(1)如果返回是list类型(不用提前知道大小)

Solution2：排序后 Two pointers

Time Complexity: O(nlogn) Space Complexity: O(N)
也可以是extraO(1)如果返回是list类型(不用提前知道大小)

Solution3：Binary Search?

思路： 如何handle duplicate？
Time Complexity: O(nlogn) Space Complexity: O(N)
也可以是extraO(1)如果返回是list类型(不用提前知道大小)

What if the given array is already sorted? How would you optimize your algorithm? [Use two pointers]
What if nums1's size is small compared to nums2's size? Which algorithm is better? [hashmap, (or binary search)]
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?
[hashmap saves nums1]

Solution Code：

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        HashMap map = new HashMap();
        ArrayList result = new ArrayList();
        for(int i = 0; i < nums1.length; i++) {
            if(map.containsKey(nums1[i])) map.put(nums1[i], map.get(nums1[i]) + 1);
            else map.put(nums1[i], 1);
        }
    
        for(int i = 0; i < nums2.length; i++) {
            if(map.containsKey(nums2[i]) && map.get(nums2[i]) > 0) {
                result.add(nums2[i]);
                map.put(nums2[i], map.get(nums2[i]) - 1);
            }
        }
    
       int[] r = new int[result.size()];
       for(int i = 0; i < result.size(); i++) {
           r[i] = result.get(i);
       }
       return r;
    }
}