Leetcode：350. Intersection of Two Arrays II 求兩個數組的交集

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?
相似的一個題

這個題要求求出交集，且重複的數字需要輸出，和349的最大的差別就是這個。顯然我們現在不能利用Set集合的特性。
思路：創建Map集合
1.把nums1加入map集合，注意：第一次加入直接鍵爲當前nums[i],值爲1，否則，若已經存在的話，則覆蓋掉原來的值，值加1.
2.遍歷nums2，如果包含那個鍵，加入result集合，讓map的該鍵的值減一，直到map.get(nums2[i]) = 0)就說明結束

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        ArrayList<Integer> result = new ArrayList<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);
        }

        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;
    }
}