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 。
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.
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
Java:
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
Map<Integer, Integer> map = new HashMap<>();
int count = 0;
for (int i = 0; i < A.length; i++)
for (int j = 0; j < B.length; j++) {
int sum = A[i] + B[j];
map.put(sum, map.getOrDefault(sum, 0) + 1); // key 为两个数组中元素组合值之和, value 为两数和的值出现的次数
}
for (int i = 0; i < C.length; i++)
for (int j = 0; j < D.length; j++)
count += map.getOrDefault(-C[i] - D[j], 0); // 找到满足条件的 key , 总次数与对应 value 值累加 (因为value 代表 A, B 数组中符合条件的组合的次数)
return count;
}
}
Python:
class Solution:
def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
count, hash_map = 0, dict()
for a in A:
for b in B:
sum_ab = a+b
hash_map.setdefault(sum_ab, 0)
hash_map[sum_ab] += 1 # key 为两个数组中元素组合值之和, value 为两数和的值出现的次数
for c in C:
for d in D:
count += hash_map.get(-c-d, 0) # 找到满足条件的 key , 总次数与对应 value 值累加 (因为value 代表 A, B 数组中符合条件的组合的次数)
return count