Description

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

Solution

HashMap, time O(n^2), space O(n)

用了二分、HashMap等解法紛紛TLE，后來發(fā)現(xiàn)必須把時間降到O(n^2)才能過。這就要求HashMap中保存a + b，而非a。

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        Map<Integer, Integer> map = new HashMap<>();
        
        for (int a : A) {
            for (int b : B) {
                int sum = a + b;
                map.put(sum, map.getOrDefault(sum, 0) + 1);
            }
        }
        
        int count = 0;
        
        for (int c : C) {
            for (int d : D) {
                int target = -c - d;
                count += map.getOrDefault(target, 0);
            } 
        }
        
        return count;
    }
}