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:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

 

这道题是之前那道的延伸，让我们在四个数组中各取一个数字，使其和为0。那么坠傻的方法就是遍历所有的情况，时间复杂度为O(n⁴)。但是我们想想既然那道都能将时间复杂度缩小一倍，那么这道题我们使用哈希表是否也能将时间复杂度降到O(n²)呢？答案是肯定的，我们如果把A和B的两两之和都求出来，在哈希表中建立两数之和跟其出现次数之间的映射，那么我们再遍历C和D中任意两个数之和，我们只要看哈希表存不存在这两数之和的相反数就行了，参见代码如下：

 

解法一：

class Solution {public:    int fourSumCount(vector
     
      & A, vector
      
       & B, vector
       
        & C, vector
        
         & D) {        int res = 0;        unordered_map
         
           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;    }};
         
        
       
      
     

 

这种方法用了两个哈希表分别记录AB和CB的两两之和出现次数，然后遍历其中一个哈希表，并在另一个哈希表中找和的相反数出现的次数，参见代码如下：

 

解法二：

class Solution {public:    int fourSumCount(vector
     
      & A, vector
      
       & B, vector
       
        & C, vector
        
         & D) {        int res = 0, n = A.size();        unordered_map
         
           m1, m2;        for (int i = 0; i < n; ++i) {            for (int j = 0; j < n; ++j) {                ++m1[A[i] + B[j]];                ++m2[C[i] + D[j]];            }        }        for (auto a : m1) res += a.second * m2[-a.first];        return res;    }};
         
        
       
      
     

 

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