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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 = 0 2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
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最开始想的是固定前三个数，然后看第四个数满足条件否
显然，超时了
而且，这时候才意识到，list 判断in的时候，是o(n)的
然后就想，A+B放一起，也就500*500个，然后再去判断CD的
嗯。。。就可以了
class Solution(object):def fourSumCount(self, A, B, C, D):res2 = {}res = 0for i in C:for j in D:res2[i+j] = res2.get(i+j,0) + 1for i in A:for j in B:res += res2.get(-i-j,0)return res
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