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: 

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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

思路

由于结果不能重复，我们就以空间换时间，经历多层哈希表得到最终的结果。

把数组A、B、C、D里面的数字出现频率统计一番，接下来把A+B，C+D当中出现过的数字也进行统计。最后遍历统计后的AB, CD两个哈希表，得到结果。

时间复杂度由naive的O(n^4) 降低到最好情况为O(1), 平均下来应也有O(n^2)

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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
        """
        countA = {}
        countB = {}
        countC = {}
        countD = {}
        ans = 0
        for i in A:
            if (countA.has_key(i)):
                countA[i] += 1
            else:
                countA[i] = 1
        for i in B:
            if (countB.has_key(i)):
                countB[i] += 1
            else:
                countB[i] = 1
        for i in C:
            if (countC.has_key(i)):
                countC[i] += 1
            else:
                countC[i] = 1
        for i in D:
            if (countD.has_key(i)):
                countD[i] += 1
            else:
                countD[i] = 1
        
        countAB = {}
        
        for i in countA.keys():
            for j in countB.keys():
                if (countAB.has_key(i+j)):
                    countAB[i+j] += countA[i] * countB[j]
                else:
                    countAB[i+j] = countA[i] * countB[j]
        countCD = {}
                
        for i in countC.keys():
            for j in countD.keys():
                if (countCD.has_key(i+j)):
                    countCD[i+j] += countC[i] * countD[j]
                else:
                    countCD[i+j] = countC[i] * countD[j]
                    
        for i in countAB.keys():
            if countCD.has_key(-i):
                ans += countAB[i] * countCD[-i]
        
        return ans

实际上直接这样就可以AC了……

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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
        """
        ans = 0
        
        countAB = {}
        
        for i in A:
            for j in B:
                if (countAB.has_key(i+j)):
                    countAB[i+j] += 1
                else:
                    countAB[i+j] = 1
        countCD = {}
                
        for i in C:
            for j in D:
                if (countCD.has_key(i+j)):
                    countCD[i+j] += 1
                else:
                    countCD[i+j] = 1
                    
        for i in countAB.keys():
            if countCD.has_key(-i):
                ans += countAB[i] * countCD[-i]
        
        return ans