4Sum II Solutions in Java

/// https://leetcode.com/problems/4sum-ii/description/
/// Author : liuyubobobo
/// Time   : 2017-11-15

import java.util.HashMap;

/// Using Hash Map
/// Time Complexity: O(n^2)
/// Space Complexity: O(n^2)
public class Solution1 {

    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {

        if(A == null || B == null || C == null || D == null)
            throw new IllegalArgumentException("Illegal argument");

        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i = 0 ; i < C.length ; i ++)
            for(int j = 0 ; j < D.length ; j ++){
                int sum = C[i] + D[j];
                if(map.containsKey(sum))
                    map.put(sum, map.get(sum) + 1);
                else
                    map.put(sum, 1);
            }

        int res = 0;
        for(int i = 0 ; i < A.length ; i ++)
            for(int j = 0 ; j < B.length ; j ++)
                if(map.containsKey(-A[i]-B[j]))
                    res += map.get(-A[i]-B[j]);

        return res;
    }

    public static void main(String[] args) {

        int[] a = {1, 2};
        int[] b = {-2, -1};
        int[] c = {-1, 2};
        int[] d = {0, 2};
        System.out.println((new Solution1()).fourSumCount(a, b, c, d));
    }
}

/// https://leetcode.com/problems/4sum-ii/description/
/// Author : liuyubobobo
/// Time   : 2017-11-15

import java.util.HashMap;

/// Another Way to use Hash Map
/// Time Complexity: O(n^2)
/// Space Complexity: O(n^2)
public class Solution2 {

    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {

        if(A == null || B == null || C == null || D == null)
            throw new IllegalArgumentException("Illegal argument");

        HashMap<Integer, Integer> mapAB = 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(mapAB.containsKey(sum))
                    mapAB.put(sum, mapAB.get(sum) + 1);
                else
                    mapAB.put(sum, 1);
            }

        HashMap<Integer, Integer> mapCD = new HashMap<Integer, Integer>();
        for(int i = 0 ; i < C.length ; i ++)
            for(int j = 0 ; j < D.length ; j ++){
                int sum = C[i] + D[j];
                if(mapCD.containsKey(sum))
                    mapCD.put(sum, mapCD.get(sum) + 1);
                else
                    mapCD.put(sum, 1);
            }

        int res = 0;
        for(Integer sumab: mapAB.keySet()){
            if(mapCD.containsKey(-sumab))
                res += mapAB.get(sumab) * mapCD.get(-sumab);
        }

        return res;
    }

    public static void main(String[] args) {

        int[] a = {1, 2};
        int[] b = {-2, -1};
        int[] c = {-1, 2};
        int[] d = {0, 2};
        System.out.println((new Solution2()).fourSumCount(a, b, c, d));
    }
}