leetcode 454. 4Sum II

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

想法就是分成两组，A+B的和以及C+D的和，使用Map保存和的频数，计算两组和的和为0时的频数乘积

public class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        Map<Integer, Integer> m1 = new HashMap<>();
        Map<Integer, Integer> m2 = new HashMap<>();
        int ans = 0;
        for(int i=0; i<A.length; i++){
            for(int j=0; j<A.length; j++)
                m1.put(A[i]+B[j], m1.getOrDefault(A[i]+B[j], 0)+1);
        }
        for(int i=0; i<A.length; i++){
            for(int j=0; j<A.length; j++)
                m2.put(C[i]+D[j], m2.getOrDefault(C[i]+D[j], 0)+1);
        }
        
        for(int key : m1.keySet()){
            if(m2.containsKey(-key))
                ans += m1.get(key) * m2.get(-key); 
        }
        return ans;
    }
}

用一个Map也可以，在算C+D的时候就可以进行判断了

public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    Map<Integer, Integer> map = new HashMap<>();
    
    for(int i=0; i<C.length; i++) {
        for(int j=0; j<D.length; j++) {
            int sum = C[i] + D[j];
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
    }
    
    int res=0;
    for(int i=0; i<A.length; i++) {
        for(int j=0; j<B.length; j++) {
            res += map.getOrDefault(-1 * (A[i]+B[j]), 0);
        }
    }
    
    return res;
}