LeetCode 454. 四数相加 II 4Sum II

4-5 灵活选择键值 4Sum II

题目: LeetCode 454. 四数相加 II

给定四个包含整数的数组列表 A , B , C , D ,计算有多少个元组 (i, j, k, l) ，使得 A[i] + B[j] + C[k] + D[l] = 0。

为了使问题简单化，所有的 A, B, C, D 具有相同的长度 N，且 0 ≤ N ≤ 500 。所有整数的范围在 -228 到 228 - 1 之间，最终结果不会超过 231 - 1 。

例如:

输入:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

输出:
2

解释:
两个元组如下:

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

import java.util.HashMap;

// 454. 4Sum II
// https://leetcode.com/problems/4sum-ii/description/
// 时间复杂度: O(n^2)
// 空间复杂度: 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));
    }
}

import java.util.HashMap;

// 454. 4Sum II
// https://leetcode.com/problems/4sum-ii/description/
// 时间复杂度: O(n^2)
// 空间复杂度: 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));
    }
}