LeetCode link:[454. 4Sum II](https://leetcode.com/problems/problem), difficulty:**Medium**.

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

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

**Input**:`nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]`

1. Because the final requirement is to take one number from each group of numbers, the four groups of numbers can be split into**two groups of two**.

2. Count the number of each`sum`. Use`Map` to store,`key` is`sum`,`value` is`count`.

3. Iterate over`nums3` and`nums4`, if`-(num3 + num4)` exists in`keys` of`Map`, then`count` is included in the total.

1. Count the number of each`sum`. Use`Map` to store,`key` is`sum`,`value` is`count`.

num_to_count= defaultdict(int)

for num1in nums1:

for num2in nums2:

num_to_count[num1+ num2]+=1

2. Iterate over`nums3` and`nums4`, if`-(num3 + num4)` exists in`keys` of`Map`, then`count` is included in the total.

result=0

for num3in nums3:

for num4in nums4:

result+= num_to_count[-(num3+ num4)]

return result

* Time:`O(n * n)`.

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：

**输入**:`nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]`

3. 遍历`nums3`和`nums4`，如果`-(num3 + num4)`存在于`Map`的`keys`中，则`count`计入总数。

1. 统计出每个`和`值对应的个数。使用`Map`储存，`key`为`和`，`value`为`count`。

num_to_count= defaultdict(int)

for num1in nums1:

for num2in nums2:

num_to_count[num1+ num2]+=1

2. 遍历`nums3`和`nums4`，如果`-(num3 + num4)`存在于`Map`的`keys`中，则`count`计入总数。

result=0

for num3in nums3:

for num4in nums4:

result+= num_to_count[-(num3+ num4)]

return result

* 时间:`O(n * n)`.