[LeetCode] 454. 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 -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

四数相加II。

题意是给定四个包含整数的数组列表 A , B , C , D ,计算有多少个元组 (i, j, k, l) ，使得 A[i] + B[j] + C[k] + D[l] = 0。为了使问题简单化，所有的 A, B, C, D 具有相同的长度 N，且 0 ≤ N ≤ 500 。所有整数的范围在 -2^28 到 2^28 - 1 之间，最终结果不会超过 2^31 - 1。

思路是hashmap。将A和B分为一组，C和D分为一组。因为求的是A + B + C + D = 0，所以能得出结论A + B = -C - D。按照这个思路，求出A + B的和的所有可能，存在hashmap里面，key是A + B的和，value是出现次数。所以当算完A + B的所有可能之后，再遍历C和D，找-C - D是否在hashmap里出现过，若出现过，则直接将出现次数加到结果中。

时间O(n^2)

空间O(n)

Java实现

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        HashMap<Integer, Integer> map = new HashMap<>();
        int res = 0;
        for (int a : A) {
            for (int b : B) {
                int sum = a + b;
                map.put(sum, map.getOrDefault(sum, 0) + 1);
            }
        }
        for (int c : C) {
            for (int d : D) {
                int sum = -c - d;
                res += map.getOrDefault(sum, 0);
            }
        }
        return res;
    }
}

JavaScript实现

/**
 * @param {number[]} A
 * @param {number[]} B
 * @param {number[]} C
 * @param {number[]} D
 * @return {number}
 */
var fourSumCount = function (A, B, C, D) {
    let map = new Map(),
        ans = 0,
        n = A.length;

    for (let i = 0; i < n; i++) {
        let a = A[i];
        for (let j = 0; j < n; j++) {
            let b = B[j];
            if (!map.has(a + b)) {
                map.set(a + b, 1);
            } else {
                map.set(a + b, map.get(a + b) + 1);
            }
        }
    }

    for (let k = 0; k < n; k++) {
        let c = C[k];
        for (let l = 0; l < n; l++) {
            let d = D[l];
            let sum = -(c + d);
            if (map.has(sum)) {
                ans += map.get(sum);
            }
        }
    }
    return ans;
};

LeetCode 题目总结