454. 4Sum II
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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
一开始想了一个O(n^3), space O(N)的做法,后来发现还可以优化
Solution: time O(N^2), space O(N^2)
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int i = 0; i < A.length; i++) { for (int j = 0; j < B.length; j++) { map.put(A[i] + B[j], map.getOrDefault(A[i] + B[j], 0) + 1); } } int res = 0; for (int i = 0; i < C.length; i++) { for (int j = 0; j < D.length; j++) { res += map.getOrDefault(-(C[i] + D[j]), 0); } } return res; }
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