454. 4Sum II

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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:

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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

解法1: O(N^2)

就是two sum的解法
C++

1

Java

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public class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        // N^3logN
        if (A.length == 0) {
            return 0;
        }
        int count = 0;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        int n = A.length;
        // O(N^2)
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int sum = A[i] + B[j];
                map.put(sum, map.getOrDefault(sum, 0) + 1);
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int sum = C[i] + D[j];
                if (map.containsKey(0 - sum)) {
                    count += map.get(0 - sum);
                }
            }
        }
        return count;
    }
}