Leetcode：350. Intersection of Two Arrays II 求两个数组的交集

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?
相似的一个题

这个题要求求出交集，且重复的数字需要输出，和349的最大的差别就是这个。显然我们现在不能利用Set集合的特性。
思路：创建Map集合
1.把nums1加入map集合，注意：第一次加入直接键为当前nums[i],值为1，否则，若已经存在的话，则覆盖掉原来的值，值加1.
2.遍历nums2，如果包含那个键，加入result集合，让map的该键的值减一，直到map.get(nums2[i]) = 0)就说明结束

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;    }}