LeetCode 350. Intersection of Two Arrays II

原题链接在这里：https://leetcode.com/problems/intersection-of-two-arrays-ii/

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?

题解：

可以使用双指针.

Time Complexity: O(nlogn). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int[] intersect(int[] nums1, int[] nums2) {
 3         Arrays.sort(nums1);
 4         Arrays.sort(nums2);
 5         int i = 0;
 6         int j = 0;
 7         List<Integer> res = new ArrayList<Integer>();
 8         while(i<nums1.length && j<nums2.length){
 9             if(nums1[i] < nums2[j]){
10                 i++;
11             }else if(nums1[i] > nums2[j]){
12                 j++;
13             }else{
14                 res.add(nums1[i]);
15                 i++;
16                 j++;
17             }
18         }
19         
20         int [] resArr = new int[res.size()];
21         int k = 0;
22         for(int num : res){
23             resArr[k++] = num;
24         }
25         return resArr;
26     }
27 }

Could use HashMap as well.

Time Complexity: O(m + n).

Space: O(Math.min(m, n)).

AC Java:

 1 class Solution {
 2     public int[] intersect(int[] nums1, int[] nums2) {
 3         if(nums1 == null || nums2 == null){
 4             return new int[0];
 5         }
 6         
 7         HashMap<Integer, Integer> hm = new HashMap<>();
 8         for(int num : nums1){
 9             hm.put(num, hm.getOrDefault(num, 0) + 1);
10         }
11         
12         List<Integer> res = new ArrayList<>();
13         for(int num : nums2){
14             if(hm.containsKey(num)){
15                 res.add(num);
16                 if(hm.get(num) == 1){
17                     hm.remove(num);
18                 }else{
19                     hm.put(num, hm.get(num) - 1);
20                 }
21             }
22         }
23         
24         int [] resArr = new int[res.size()];
25         int i = 0;
26         for(int num : res){
27             resArr[i++] = num;
28         }
29         
30         return resArr;
31     }
32 }

Follow up 2 nums1 length is smaller. 用双指针先sort两个array明显没有利用到num1.length小的特性. 若是用HashMap来记录num1每个element出现频率再iterate nums2, 那么Time Complexity: O(m + n), m = nums1.length, n = num2.length. Space: O(m).

或者sort nums1 再对每一个num2的element在 sorted nums1上做 binary search. Time Complexity: O(mlogm + nlogm). Space: O(1).

由此可见，当m很小时，用HashMap和binary search就是time和space的trade off.

Follow up 3 nums2 is sorted but too big for memory. I/O的操作很贵，所以首先想到的是避免I/O的次数。

若是nums1可以全部load到memory上, 先sort nums1再把nums2从小到大分开load到memory来.

　　if load进来这一段最大值, 也就是最后一个值<nums1[0] 或者 load进来这一段最小值, 也就是第一个值>nums1[nums1.length-1]可以直接跳过, load下一段. 　　else load进来这一段和 sorted nums1做双指针.

若是nums1也太大了，就先external sort nums1, 在分开load进来nums1一段和nums2一段做双指针.

类似Intersection of Two Arrays.