题目信息

• 链接: 4sum-ii

• Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:

  • 0 <= i, j, k, l < n
  • nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

  Example 1:

  Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
  Output: 2
  Explanation:
  The two tuples are:
  1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
  2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
  
  

  Example 2:

  Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
  Output: 1
  
  

  Constraints:

  • n == nums1.length
  • n == nums2.length
  • n == nums3.length
  • n == nums4.length
  • 1 <= n <= 200
  • -2<sup>28</sup> <= nums1[i], nums2[i], nums3[i], nums4[i] <= 2<sup>28</sup>

• 描述:

  • 给你四个整数数组 nums1、nums2、nums3 和 nums4 ，数组长度都是 n ，请你计算有多少个元组 (i, j, k, l) 能满足：
  • 0 <= i, j, k, l < n
    nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

解题思路

• 暴力法循环4个数组$O(n^4)$，使用哈希表可以优化到$O(n^2)$
• 将4个数组两两分组，遍历第一组所有的a+b和，并构建哈希表key为a+b，value为a+b出现的次数
• 遍历第二组c+d，在哈希表查找0-c-d，获得其出现次数value，cnt+=value

代码

  class Solution:
	  def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
		  s_dict = {}
		  result = 0
		  for a in nums1:
			  for b in nums2:
					  s_dict[a+b] = s_dict.get(a+b, 0) + 1
		  for c in nums3:
			  for d in nums4:
				  value = s_dict.get(0-c-d)
				  if value:
					  result += value
		  return result