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454. 四数相加 II

English Version

题目描述

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 -2²⁸ 到 2²⁸ - 1 之间，最终结果不会超过 2³¹ - 1 。

例如:

输入:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

输出:
2

解释:
两个元组如下:
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

解法

Python3

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        counter = Counter()
        for a in nums1:
            for b in nums2:
                counter[a + b] += 1
        ans = 0
        for c in nums3:
            for d in nums4:
                ans += counter[-(c + d)]
        return ans

Java

class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int a : nums1) {
            for (int b : nums2) {
                counter.put(a + b, counter.getOrDefault(a + b, 0) + 1);
            }
        }
        int ans = 0;
        for (int c : nums3) {
            for (int d : nums4) {
                ans += counter.getOrDefault(-(c + d), 0);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        unordered_map<int, int> counter;
        for (int a : nums1)
            for (int b : nums2)
                ++counter[a + b];
        int ans = 0;
        for (int c : nums3)
            for (int d : nums4)
                ans += counter[-(c + d)];
        return ans;
    }
};

Go

func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {
	counter := make(map[int]int)
	for _, a := range nums1 {
		for _, b := range nums2 {
			counter[a+b]++
		}
	}
	ans := 0
	for _, c := range nums3 {
		for _, d := range nums4 {
			ans += counter[-(c + d)]
		}
	}
	return ans
}

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