349. Intersection of Two Arrays --双指针、字典、二分查找、集合

349. Intersection of Two Arrays

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

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Solution 1:

use set operation in python, one-line solution.

class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
return list(set(nums1) & set(nums2))
Solution 2:

brute-force searching, search each element of the first list in the second list. (to be more efficient, you can sort the second list and use binary search to accelerate)

class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
res = []
for i in nums1:
if i not in res and i in nums2:
res.append(i)

return res
Solution 3:

use dict/hashmap to record all nums appeared in the first list, and then check if there are nums in the second list have appeared in the map.

class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
res = []
map = {}
for i in nums1:
map[i] = map[i]+1 if i in map else 1
for j in nums2:
if j in map and map[j] > 0:
res.append(j)
map[j] = 0

return res
Solution 4:

sort the two list, and use two pointer to search in the lists to find common elements.

class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
res = []
nums1.sort()
nums2.sort()
i = j = 0
while (i < len(nums1) and j < len(nums2)):
if nums1[i] > nums2[j]:
j += 1
elif nums1[i] < nums2[j]:
i += 1
else:
if not (len(res) and nums1[i] == res[len(res)-1]):
res.append(nums1[i])
i += 1
j += 1

return res