### 4Sum II

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 -2^28 to 2^28 - 1 and the result is guaranteed to be at most 2^31 - 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 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

### Analyze

/** * @param {number[]} A * @param {number[]} B * @param {number[]} C * @param {number[]} D * @return {number} */var fourSumCount = function(A, B, C, D) {  const tmpMap = new Map()  for (let ic = 0; ic < C.length; ic++) {    for (let id = 0; id < D.length; id++) {      const sumC_D = C[ic] + D[id]      const hasSumC_D = tmpMap.has(sumC_D)      if (hasSumC_D) {        tmpMap.set(sumC_D, tmpMap.get(sumC_D) + 1)      } else {        tmpMap.set(sumC_D, 1)      }    }  }
let count = 0
for (let ia = 0; ia < A.length; ia++) {    for (let ib = 0; ib < B.length; ib++) {      const sumA_B = A[ia] + B[ib]      tmpMap.has(-sumA_B) && (count = count + tmpMap.get(-sumA_B))    }  }
return count}

### Sister Title

18(题目相似, 但是因为传入参数不同导致解法不同)