[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

 思路: 

1.把四個 list 分成兩組 A, B 和 C, D

2.用hash table 存入 A,B 的所有排列組合, 即 A[0]+B[0], A[0]+B[1], ...

3.如果 -(c+d) 可以在hash table 裡找到 表示 合為 0

 

  class Solution(object):
    def fourSumCount(self, A, B, C, D):
        """
        :type A: List[int]
        :type B: List[int]
        :type C: List[int]
        :type D: List[int]
        :rtype: int
        """
        dic, res = {}, 0
        for a in A:
            for b in B:
                if a+b not in dic: dic[a+b] = 1
                else: dic[a+b] += 1
       
        for c in C:
            for d in D:
                if -(c+d) in dic:
                    res += dic[-(c+d)]
        return res