454. 4Sum II

Posted 2020-10-20 Long Long Journey

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。

/**
 * @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);

来自为知笔记(Wiz)