数组的交集——倒排索引查询，因为不重复，还是可以采用计数的思路做，外部排序，二路归并的做法也不错

793. 多个数组的交集

中文

English

给出多个数组，求它们的交集。输出他们交集的大小。

样例

样例 1:

	输入:  [[1,2,3],[3,4,5],[3,9,10]]
	输出:  1
	
	解释:
	只有3出现在三个数组中。

样例 2:

	输入: [[1,2,3,4],[1,2,5,6,7][9,10,1,5,2,3]]
	输出:  2
	
	解释:
	交集是[1,2].

注意事项

• 输入的所有数组元素总数不超过500000。
• 题目数据每个数组里的元素没有重复。

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import heapq     class Solution:     """     @param arrs: the arrays     @return: the number of the intersection of the arrays     """     def intersectionOfArrays(self, arrs):         # write your code here         q = []                   for i, arr in enumerate(arrs):             if arr:                 arr.sort()                 heapq.heappush(q, (arrs[i][0], i, 0))                   ans = 0         prev = float('inf')         cnt = 1         while q:             val, i, j = heapq.heappop(q)                           if val == prev:                 cnt += 1                 if cnt == len(arrs):                     ans += 1             else:                 prev = val                 cnt = 1                                   if j+1 < len(arrs[i]):                 heapq.heappush(q, (arrs[i][j+1], i, j+1))                   return ans

 此外，还可以使用 二路归并的做法：

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import heapq     class Solution:     """     @param arrs: the arrays     @return: the number of the intersection of the arrays     """     def intersectionOfArrays(self, arrs):         # write your code here         def intersect2(arr1, arr2):             return list(set(arr1) & set(arr2))                   def intersect(arr, l, r):             if l == r:                 return arr[l]                           if l > r:                 return []                               mid = (l+r) >> 1             arr1 = intersect(arr, l, mid)             arr2 = intersect(arr, mid+1, r)             return intersect2(arr1, arr2)                   return len(intersect(arrs, 0, len(arrs)-1))