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Given two arrays, write a function to compute their intersection.

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2] Example 2: Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9] 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?

我的想法

一开始题意理解错了，以为是substring。这里是求交集，对顺序是没有要求的

class Solution {
       public int[] intersect(int[] nums1, int[] nums2) {
           HashMap
   
     map = new HashMap<>();        List
    
      res = new ArrayList
     
      ();        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])) {
                   res.add(nums2[i]);                if(map.get(nums2[i]) > 1) {
                       map.put(nums2[i], map.get(nums2[i]) - 1);                } else {
                       map.remove(nums2[i]);                }            }        }        int[] retn = new int[res.size()];        for(int i = 0; i < res.size(); i++) {
               retn[i] = res.get(i);        }        return retn;    }}
     
    
   

解答

follow up:

1: 用双指针，比hashmap省空间

2:

用HashMap，把小的存在hashmap里

3:

If only nums2 cannot fit in memory, put all elements of nums1 into a HashMap, read chunks of array that fit into the memory, and record the intersections.

If both nums1 and nums2 are so huge that neither fit into the memory, sort them individually (external sort), then read 2 elements from each array at a time in memory, record intersections.

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