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
能够优化到O(n2),但是可以只用一个map,如下,我本来的想法是用两个map,但这样就浪费了空间。
Runtime: 155 ms, faster than 67.41% of Java online submissions for 4Sum II.
class Solution { private Map<Integer, Integer> mp1 = new HashMap<>(); public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { for(int i = 0; i<A.length; i++){ for(int j=0; j<B.length; j++){ mp1.put(A[i] + B[j], mp1.getOrDefault(A[i]+B[j],0) + 1); } } int ret = 0; for(int i = 0; i<C.length; i++){ for(int j=0; j<D.length; j++){ ret += mp1.getOrDefault(-1*(C[i]+D[j]), 0); } } return ret; } }