[LeetCode] 454. 4Sum II 四数之和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

在给定的4个数组中各取1个数字，使4个数的和为0，返回找到的次数。

如果用暴力搜索Bruce Force，那么Time: O(n^4)，用时太长。

好的解法是用Hash map，建立一个map，将A，B中每两个数的和作为key，次数作为value写入map，然后查找C，D中每两个数和的相反数是否在map中，如果在，就把 map 中记录的次数累加到结果中。

Java：

public class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int res = 0;
        HashMap<Integer,Integer> ab = new HashMap<Integer,Integer>();

        for(int a:A){
            for(int b:B){
                ab.put(a+b,ab.getOrDefault(a+b,0) + 1);
            }
        }
        for(int c:C){
            for(int d:D){
                int part2 = c + d;
                int part1 = - part2;
                res += ab.getOrDefault(part1,0);
            }
        }
        return res;
    }
}

Python:

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        ans = 0
        cnt = collections.defaultdict(int)
        for a in A:
            for b in B:
                cnt[a + b] += 1
        for c in C:
            for d in D:
                ans += cnt[-(c + d)]
        return ans

Python: 这个写法牛逼到了只用两行就搞定了

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        A_B_sum = collections.Counter(a+b for a in A for b in B)
        return sum(A_B_sum[-c-d] for c in C for d in D)

C++:

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int res = 0;
        unordered_map<int, int> m;
        for (int i = 0; i < A.size(); ++i) {
            for (int j = 0; j < B.size(); ++j) {
                ++m[A[i] + B[j]];
            }
        }
        for (int i = 0; i < C.size(); ++i) {
            for (int j = 0; j < D.size(); ++j) {
                int target = -1 * (C[i] + D[j]);
                res += m[target];
            }
        }
        return res;
    }
};

　　　　

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