LeetCode: 4Sum II

Time Complexity: O(n^2)

public class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int len = A.length;
        int ans = 0;
        Map<Integer, Integer> countAB = new HashMap<Integer, Integer>();
        Map<Integer, Integer> countCD = new HashMap<Integer, Integer>();
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < len; j++) {
                Integer g = new Integer(A[i] + B[j]);
                if (countAB.containsKey(g)) {
                    int v = countAB.get(g);
                    countAB.put(g, ++v);
                }
                else countAB.put(g, 1);
                g = new Integer(C[i] + D[j]);
                if (countCD.containsKey(g)) {
                    int v = countCD.get(g);
                    countCD.put(g,  ++v);
                }
                else countCD.put(g, 1);
            }
        }
        for (Integer key : countAB.keySet()) {
            if (countCD.containsKey(-key)) {
                ans += countAB.get(key) * countCD.get(-key);
            }
        }
        return ans;
    }
}