leetcode 350 Intersection of Two Arrays II (两个数组的交集 II) python3 多种思路，最简代码

所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。

class Solution:
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        
 # 巧用python库中的Counter()来构造字典
        # Approach one： 只构造一个字典,nums2比较大时，节省一次构造的时间
        # from collections import Counter
        # result = []   
        # dic1 = Counter(nums1)
        # for num in nums2:
        #     if num in dic1.keys():
        #         result.append(num)
        #         dic1[num] -= 1        # 通过字典操作实现检测个数的要求
        #         if dic1[num] == 0:
        #             dic1.pop(num)            
        # return result
        
        
         # Approach two： 构造两个字典则不用频繁的字典操作
        # from collections import Counter
#         result = []   
#         dic1,dic2 = Counter(nums1),Counter(nums2)
#         for num in dic2.keys():
#             if num in dic1.keys():
#                 repeat = min(dic1.get(num),dic2.get(num))
#                 for i in range(repeat):
#                     result.append(num)
#         return result



        # Approach #3   直接对字典操作，注意dic.elements()  返回的是一个迭代器。需要list()才能正常输出。   
        # 这里用values()没办法输出重复的值。
        return list((collections.Counter(nums1)&collections.Counter(nums2)).elements())
        
        
        
        
        # Approach #4  two pointers solution
        # nums1.sort(), nums2.sort()  # Make sure it is sorted, doesn't count in time.
        # res = []
        # it1, it2 = 0, 0
        # while it1 < len(nums1) and it2 < len(nums2):
        #     if nums1[it1] < nums2[it2]:
        #         it1 += 1
        #     elif nums1[it1] > nums2[it2]:
        #         it2 += 1
        #     else:
        #         res += nums1[it1],
        #         it1 += 1
        #         it2 += 1
        # return res
        
        
        
        
        
        # Approach five
        # res  = []
        # dic1 = {}
        # for i in nums1:
        #     dic1[i] = dic1.get(i,0) + 1
        # dic2 = {}
        # for i in nums2:
        #     dic2[i] = dic2.get(i,0) + 1
        # for k,v in dic1.items():
        #     if v < dic2.get(k,0):
        #         res += [k]*v
        #     else:
        #         res += [k]*dic2.get(k,0)
        # return res

        
        

所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。