Intersection of Two Arrays 2

原题地址：https://leetcode.com/problemset/algorithms/

题目：

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?

解题思路：
这道题解题思路可以用map，也可以用先sort后找duplicate的办法。如果是两个array的size都比较大，那么可以先把他们用external sort先分别排序，然后分别拿出一个chunk来找重复部分。

代码：

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