leetcode 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?

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        # use sort, or set(Counter)
        ans = []
        cnt = collections.Counter(nums1)
        for n in nums2:
            if n in cnt and cnt[n] > 0:
                cnt[n] -= 1
                ans.append(n)
        return ans

可以更精简些：

from collections import Counter

class Solution(object):
    def intersect(self, nums1, nums2):
        c1, c2 = Counter(nums1), Counter(nums2)
        return sum([[num] * min(c1[num], c2[num]) for num in c1 & c2], [])

自己写的话：

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

        counts = {}
        res = []

        for num in nums1:
            counts[num] = counts.get(num, 0) + 1

        for num in nums2:
            if num in counts and counts[num] > 0:
                res.append(num)
                counts[num] -= 1

        return res

注意：

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        # use sort, or set(Counter)
        return list(set(nums1) & set(nums2))是不可以的，因为set里是没有重复数据的：
Input: [1,2,2,1] [2,2]
Output: [2]
Expected: [2,2]

另外就是使用sort，归并思路：

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        # use sort, or set(Counter)
        nums1.sort()
        nums2.sort()
        i = j = 0
        ans = []
        l1, l2 = len(nums1), len(nums2)
        while i < l1 and j < l2:
            if nums1[i] == nums2[j]:
                ans.append(nums1[i])
                i += 1
                j += 1
            elif nums1[i] < nums2[j]:
                i += 1
            else:
                j += 1
        return ans

 还有使用binsearch来避免归并思路的。