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

posted @ 2017-03-21 02:36  CodesKiller  阅读(142)  评论(0编辑  收藏  举报