Difficulty:easy
More:【目录】LeetCode Java实现
Description
https://leetcode.com/problems/intersection-of-two-arrays-ii/
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?
Intuition
Method 1: HashMap
Method 2: Sorted arrays and two pointers
Solution
//Method 1: HashMap
public int[] intersect(int[] nums1, int[] nums2) {
HashMap<Integer,Integer> map = new HashMap<>();
ArrayList<Integer> list = new ArrayList<>();
for(int i : nums1){
if(map.containsKey(i)){
map.put(i, map.get(i)+1);
}else{
map.put(i, 1);
}
}
for(int i : nums2){
if(map.containsKey(i) && map.get(i)>0){
map.put(i, map.get(i)-1);
list.add(i);
}
}
int[] arr = new int[list.size()];
for(int n=0; n<list.size(); n++){
arr[n]=list.get(n);
}
return arr;
}
//Method 2: Sorted array & Two pointers
public int[] intersect2(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
ArrayList<Integer> list = new ArrayList<>();
int i=0, j=0;
while(i<nums1.length && j<nums2.length){
if(nums1[i]<nums2[j]){
i++;
}else if (nums1[i]>nums2[j]){
j++;
}else{
list.add(nums1[i++]);
j++;
}
}
int[] arr = new int[list.size()];
for(int n=0; n<list.size(); n++){
arr[n]=list.get(n);
}
return arr;
}
Complexity
Method 1:
Time complexity : O(N)
Space complexity : O(N)
Method 2:
Time complexity : O(NlogN)
Space complexity : O(1)
More:【目录】LeetCode Java实现