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
给定四个列表A，B，C，D的整数值，计算有多少个元组（i，j，k，l），使得A [i] + B [j] + C [k] + D [1]是零。
为了使问题更容易，所有A，B，C，D具有相同的长度N，其中0≤N≤500。所有整数在-228到228-1的范围内，结果保证最多为231 - 1。
为了使问题更容易，所有A，B，C，D具有相同的长度N，其中0≤N≤500。所有整数在-228到228-1的范围内，结果保证最多为231 - 1。
/*** @param {number[]} A* @param {number[]} B* @param {number[]} C* @param {number[]} D* @return {number}*/var fourSumCount = function (A, B, C, D) {let res = 0;let m = {};for (let i in A) {for (let j in B) {let value = A[i] + B[j];m[value] = m[value] ? ++m[value] : 1;}}for (let i in C) {for (let j in D) {let negativeValue = -1 * (C[i] + D[j]);if (m[negativeValue]) {res += m[negativeValue];}}}return res;};let A = [1, 2];let B = [-2, -1];let C = [-1, 2];let D = [0, 2];let res = fourSumCount(A, B, C, D);console.log(res);
