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?

本题和intersection of two arrays1比较类似，只不过允许出现重复的元素了，那么能否用之前的三种方法来解决这个呢？二分查找那个肯定不可以，因为不可以动态删除数组元素；数组排序应该是可以的；另外时间复杂度O（n）的那个方法也是可以的，先说时间复杂度O（n）这个，只要把hashset换成hashmap就可以了，代码如下：

 1 public class Solution {
 2     public int[] intersect(int[] nums1, int[] nums2) {
 3         Map<Integer,Integer> map = new HashMap<Integer,Integer>();
 4         List<Integer> res = new ArrayList<Integer>();
 5         for(int i:nums1){
 6             map.put(i,map.getOrDefault(i,0)+1);
 7         }
 8         for(int i:nums2){
 9             if(map.containsKey(i)&&map.get(i)>0){
10                 //find the int that appears in nums1,this method avoid remove operation
11                 map.put(i,map.get(i)-1);
12                 res.add(i);
13             }
14         }
15         int[] result = new int[res.size()];
16         int k=0;
17         for(Integer i:res){
18             result[k++] = i;
19         }
20         return result;
21     }
22 }
23 //the run time complexity costs O(max(m,n)),the space complexity O(min(m,n));

数组排序代码如下：

 1 public class Solution {
 2     public int[] intersect(int[] nums1, int[] nums2) {
 3         Map<Integer,Integer> map = new HashMap<Integer,Integer>();
 4         Arrays.sort(nums1);
 5         Arrays.sort(nums2);
 6         int count=  0;
 7         int i=0;
 8         int j=0;
 9         while(i!=nums1.length&&j!=nums2.length){
10             if(nums1[i]<nums2[j]){
11                 i++;
12             }else if(nums1[i]>nums2[j]){
13                 j++;
14             }else{
15                 count++;
16                 map.put(nums1[i],map.getOrDefault(nums1[i],0)+1);
17                 i++;
18                 j++;
19             }
20         }
21         int[] res = new int[count];
22         int v = 0;
23         for(int key:map.keySet()){
24             for(int k=0;k<map.get(key);k++){
25                 res[v++] = key;
26             }
27         }
28         return res;
29     }
30 }

本题的follow up的第三个问题有点难度，要看链接：

https://discuss.leetcode.com/topic/45992/solution-to-3rd-follow-up-question