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Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:A = [ 1, 2]B = [-2,-1]C = [-1, 2]D = [ 0, 2]Output:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

和之前的3sum有所不同，这个题的每个数在不同数组里，所以直接用4个for循环即可。

把ABCD分成两组计算，计算速度会快一些。

class Solution:    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:        dic = {}        for a in A:            for b in B :                dic[a+b]=dic.get(a+b,0)+1        count = 0                 for c in C :            for d in D :                if -c-d in dic:                    count += dic[-c-d]        return count

用collections可以写的更简单一些。

class Solution:    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:        sumab=collections.Counter(a+b for a in A for b in B)        return sum(sumab[-c-d] for c in C for d in D)

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