Leetcode_454_4Sum II

Posted 2022-11-26 皮斯特劳沃

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

思路：
（1）该题为给定数组A、B、C、D，从中分别任取一个元素，使得其和为0的组合的个数。
（2）考虑将A、B中元素的两两之和求出，将元素之和及其出现的次数存入hashmap中；再遍历C、D中任意两数之和，判断hashmap中是否存在两数之和的相反数，记录出现的次数即为所求。
（3）代码实现如下所示。希望对你有所帮助。

package leetcode;

import java.util.HashMap;

public class _4Sum_II 
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) 

        if (A == null || B == null || C == null || D == null)
            return 0;

        int sum = 0;

        HashMap<Integer, Integer> maps = new HashMap<>();

        // 将A、B各元素之和及次数存入map中
        for (int i = 0; i < A.length; i++) 
            for (int j = 0; j < A.length; j++) 

                int s = A[i] + B[j];

                if (maps.containsKey(s)) 
                    maps.put(s, maps.get(s) + 1);
                 else 
                    maps.put(s, 1);
                
            
        

        // 从map中寻找C、D之和取反的值是否存在
        for (int i = 0; i < A.length; i++) 
            for (int j = 0; j < A.length; j++) 
                int s = -1 * (C[i] + D[j]);
                if (maps.get(s) != null) 
                    sum = sum + maps.get(s);
                
            
        

        return sum;