LeetCode349 Intersection of Two Arrays
Author: Stefan Su
Create time: 2022-10-31 10:30:01
Location: New York City, NY, USA
Description Easy
Given two integer arrays nums1
and nums2
, return an array of their intersection. Each element in the result must be unique and you may return the result in any order.
Example 1
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]
Example 2
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Explanation: [4,9] is also accepted.
Constrains
1 <= nums1.length, nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 1000
Analysis
Use hash array to solve this problem. First define a hash array since the problem gives us the max size of it. And the value each number in nums1
will be the index of hash array. And value in that index will add one. If each element in nums2
which has hash[element in nums2] = 1
, then it indicates this value of index appears both in nums1
and nums2
. If hash[element in nums2] = 1
, element in nums2
appears in nums1
, which means intersection.
Solution
- unordered set version
class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
# unordered set version
return list(set(nums1) & set(nums2))
- hash table version
class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
# hash table version - based on leetcode given constrains,
# otherwise if we define a hash table too small or too large,
# that will cause conflicts or memory waste
hash = [0] * 1005
result = []
# if a number in nums1, then update the index = number to 1
for _ in nums1:
hash[_] = 1
for _ in nums2:
if hash[_] == 1:
result.append(_)
return list(set(result))
Hopefully, this blog can inspire you when solving LeetCode349. For any questions, please comment below.