我有一个问题，以找到共同的元素在两个数组，这就是不同的尺寸。

I have a problem to find common elements in two arrays and that's of different size.

取，阵列 A1 尺寸 N 和Array A2 尺寸 M 和米！= N

Take , Array A1 of size n and Array A2 of size m, and m != n

到目前为止，我已经试过遍历名单一个个元素复制到另一个列表。如果元素已经包含标记，但我知道这不是一个很好的解决方案。

So far, I've tried to iterate lists one by one and copy elements to another list. If the element already contains mark it, but I know it's not a good solution.

推荐答案

排序阵列。然后遍历它们与两个指针，始终推动所述一个指向较小的值。当它们指向相同的值，你有一个共同的价值。这将是为O（n + m），其中n和m是两个列表的大小。这就像在合并归并排序的，但如果你只产生输出时所指出的值是相等的。

Sort the arrays. Then iterate through them with two pointers, always advancing the one pointing to the smaller value. When they point to equal values, you have a common value. This will be O(n+m) where n and m are the sizes of the two lists. It's just like a merge in merge sort, but where you only produce output when the values being pointed to are equal.

def common_elements(a, b):
  a.sort()
  b.sort()
  i, j = 0, 0
  common = []
  while i < len(a) and j < len(b):
    if a[i] == b[j]:
      common.append(a[i])
      i += 1
      j += 1
    elif a[i] < b[j]:
      i += 1
    else:
      j += 1
  return common

print 'Common values:', ', '.join(map(str, common_elements([1, 2, 4, 8], [1, 4, 9])))

输出

Common values: 1, 4

如果该元素是不可比的，扔从一个列表中的元素融入到一个HashMap和检查元素对HashMap中的第二个列表。

If the elements aren't comparable, throw the elements from one list into a hashmap and check the elements in the second list against the hashmap.