[LeetCode]--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.
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?

其实跟一差不多，但是这个就是要注意短的数组不一定在后面，也可能nums1是短的那个数组。

public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        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);
        List<Integer> list = new ArrayList<Integer>();
        for (int i = 0; i < nums2.length; i++)
            if (map.containsKey(nums2[i])) {
                map.put(nums2[i], map.get(nums2[i]) - 1);
                if (map.get(nums2[i]) >= 0)
                    list.add(nums2[i]);
            }
        int[] temp = new int[list.size()];
        for (int i = 0; i < temp.length; i++) {
            temp[i] = list.get(i);
        }
        return temp;
    }