题目
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 到 2 - 1 之间,最终结果不会超过 2 - 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:
(0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
(1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
方案一(A+B = -C - D)
class Solution:
def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
# A + B = -(C + D)
N = len(A)
# d1, d2 的 key 为两数之和,值为两数的 index
d1 = {}
d2 = {}
for i in range(N):
for j in range(N):
count1 = A[i] + B[j]
count2 = C[i] + D[j]
if count1 not in d1:
d1[count1] = []
if count2 not in d2:
d2[count2] = []
d1[count1].append([i, j])
d2[count2].append([i, j])
res = 0
for key, value in d1.items():
if -key in d2:
res += len(value) * len(d2[-key])
return res
class Solution:
def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
dic = collections.Counter(a+b for a in A for b in B)
return sum(dic.get(-c-d, 0) for c in C for d in D)
参考链接
https://leetcode-cn.com/explore/orignial/card/all-about-lockup-table/239/learn-to-use-keys/998/