## Description

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.

• 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?

## Solutions

### 1. Hash Table

# Time: O(n+m)
# Space: O(n)
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
hash_map = collections.Counter(nums1)
res = []
for num in nums2:
if num in hash_map and hash_map[num] > 0:
res.append(num)
hash_map[num] -= 1
return res

# 61/61 cases passed (48 ms)
# Your runtime beats 56.08 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


### 1. What if the given array is already sorted? How would you optimize your algorithm?

如果已经排好序了，那么可以用双指针来进行操作。

class Solution(object):
def intersect(self, nums1, nums2):

nums1, nums2 = sorted(nums1), sorted(nums2)
pt1 = pt2 = 0
res = []

while True:
try:
if nums1[pt1] > nums2[pt2]:
pt2 += 1
elif nums1[pt1] < nums2[pt2]:
pt1 += 1
else:
res.append(nums1[pt1])
pt1 += 1
pt2 += 1
except IndexError:
break

return res


同第三个。

### 3. 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?

• 如果 nums1 能够放进去，那么用哈希表存下 nums1 的结果，然后分块取 nums2 的结果计算交集
• 如果两个都放不进内存，那么先分别在外部排序，然后利用两个指针的方式，每次取两个数组的部分来算交集