Leetcode350 Intersection of Two Arrays II &Leetcode 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, 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?

题意：求两个数组的交集

解法：

击败了60%+的答案

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        C = collections.Counter
        return list((C(nums1) & C(nums2)).elements()) #&代表求交集运算

这个解法的牛逼在于~惭愧，我第一次知道有Collections这个类

Counter是Collections这个类的一个子类，详情可参考 http://www.pythoner.com/205.html

例子：

import collections
nums1=[3,1,4,4,2,2,]
nums2=[3,3,5,6,2,2,2]
C = collections.Counter

C(nums1)
Out[1]: Counter({1: 1, 2: 2, 3: 1, 4: 2})

C(nums2)
Out[2]: Counter({2: 3, 3: 2, 5: 1, 6: 1})

C(nums1)&C(nums2)
Out[3]: Counter({2: 2, 3: 1})

list((C(nums1)&C(nums2)).elements())
Out[11]: [2, 2, 3]

这道题里最快的答案：

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        maps={}
        for n in nums1:
            if n in maps:
                maps[n]+=1
            else:
                maps[n]=1
        result=[]
        for i in range(len(nums2)):
            if nums2[i] in maps and maps[nums2[i]]>0:#>0其实是在取交集
                result.append(nums2[i])
                maps[nums2[i]]-=1
        return result

时间复杂度:O(n)

思路：对于nums1，把 “数字-出现次数“这个键值对保存在maps字典里

然后遍历nums2，如果nums2的某数字出现在maps1里且maps里该数字剩余的出现次数>0，就表示对于nums2里的该数字而言，它是与nums1的交集，所以放到结果result列里。然后maps[]里把该数字的剩余出现次数-1（代表该数字已经放在结果集里/代表nums2里的该数字已经被考虑过）

Leetcode349 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].

Note:

• Each element in the result must be unique.
• The result can be in any order.

		class Solution(object):

		    def intersection(self, nums1, nums2):

		        """

		        :type nums1: List[int]

		        :type nums2: List[int]

		        :rtype: List[int]

		        """

		        nums1=set(nums1)

		        nums2=set(nums2)

		        return list(nums1&nums2)

		

注意  & 其实已经是求交集了，但是对两个set求交集，结果仍是set

要求返回list， 所以list()这样强制类型转换

解法二：我自己参考上题的解法 ，不用set()函数（超级慢，only beat 5.38%的用户）

 

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

        result=[]
        for i in range(len(nums2)):
            if nums2[i] in maps  and nums2[i] not in result:
                result.append(nums2[i])
            
        return result