4sumii

problem description:

　　there is four number list named A,B,C,D; now you should out put the num of  tuples which statisfy A[i] +B[j]+C[k]+D[l] =0

i.e:

Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
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

解法：主要方法采用的是哈希表。将A和B列表中的每个值都进行两两相加，将得出的值作为哈希表的索引然后将哈希表该索引下的值加一。然后将C和D中的每个值两两相加，将所得出的值的相反数作为哈希表的索引，即可
得知相应的值有几种解法，然后进行累加。
python 实现方式：

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        """
        :type A: List[int]
        :type B: List[int]
        :type C: List[int]
        :type D: List[int]
        :rtype: int
        """
        dicts = {}
        for i in A:
            for j in B:
                sums = i+j
                if sums not in dicts:
                    dicts[sums] = 1
                else:
                    dicts[sums] += 1
        out = 0
        for k in C:
            for l in D:
                sums =-(k + l)
                if sums in dicts:
                    out += dicts[sums]
        return out
                

以上的方法虽有不错，但是还不是最精炼的python代码：

最精炼的python代码，用到了counter模块下的collection函数。它能够统计出相同数值的个数，本质上还是个哈希表

def fourSumCount(self, A, B, C, D):
    AB = collections.Counter(a+b for a in A for b in B)
    return sum(AB[-c-d] for c in C for d in D)

以上的第一个代码是个人想法，第二个精炼的代码参考自leetcode上stefanporchmann的代码