[LeetCode] 454. 4Sum II

轉自LeetCode

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

<Solution>
這題最暴力的解法，就是找出所有和，這樣時間複雜度會是 O(n^4)

使用 hash map 來做簡化到 O(n^2)

• 用兩個 hash map 來存 A+B 和 C+D 的所有情況

• 如果 A+B+C+D 可以等於 0，那就代表 A+B = -(C+D)。因此，可以用 map[-(A+B)] 來看是不是存在相對應的 C+D

code 如下(參考資料)

C++

Java