【leetcode】454. 4Sum II【M】

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:
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

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最開始想的是固定前三個數，然後看第四個數滿足條件否

顯然，超時了

而且，這時候才意識到，list 判斷in的時候，是o(n)的

然後就想，A+B放一起，也就500*500個，然後再去判斷CD的

嗯。。。就可以了

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        res2 = {}
        res = 0

        for i in C:
            for j in D:
                res2[i+j] = res2.get(i+j,0) + 1

        for i in A:
            for j in B:
                res += res2.get(-i-j,0)

        return res