454. 4Sum II [LeetCode]

Given four lists A, B, C, D of integer values, compute how many tuples(i, j, k, l)there are such thatA[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

需要记录所有重复的情况，因此需要用HashMap来计数，加O(n^2)时间复杂度

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int count = 0;
        if (A == null || B == null || C == null || D == null) {
            return count;
        }
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < A.length; i++) {
            for (int j = 0; j < B.length; j++) {
                int sum = A[i] + B[j];
                if (map.containsKey(sum)) {
                    map.put(sum, map.get(sum) + 1);
                } else {
                    map.put(sum, 1);
                }
            }
        }
        for (int i = 0; i < C.length; i++) {
            for (int j = 0; j < D.length; j++) {
                int required = -1 * (C[i] + D[j]);
                if (map.containsKey(required)) {
                    count += map.get(required);
                }
            }
        }
        return count;
    }
}