题目地址：https://leetcode.com/problems/4sum-ii/description/
题目描述
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.
Tomake 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:
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
题目大意
有四个等长的数组，分别找出四个数组中的下标，能让四个数组该位置的和是0，统计这种组合的次数。
解题方法
字典
蛮力求解肯定不行的。给出了4个数，把A,B一组，C,D一组先进行遍历求和保存到字典中，就能把复杂度从O(n4)降到O(n2)
具体的是使用Counter计算，具体的实现不难，复杂度是O(n^2)。
方法是先对A,B能组成的和进行统计，然后对C,D遍历求和并取相反数，看在A,B中出现了多少次。
代码：
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
"""
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)
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不能对两个字典进行遍历求解，那样复杂度增大，而是直接使用对一个遍历，对另个进行直接查询的方式，C++代码如下：
class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
const int N = A.size();
unordered_map<int, int> AB;
unordered_map<int, int> CD;
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
++AB[A[i] + B[j]];
++CD[C[i] + D[j]];
}
}
int res = 0;
for (auto ab : AB) {
res += ab.second * CD[-ab.first];
}
return res;
}
};
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