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
A:
-----------------用map --------------------class Solution { public: int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) { unordered_map<int,int> AB; for(auto a:A) for(auto b:B) AB[a+b]++; unordered_map<int,int> CD; for(auto c: C) for(auto d: D) CD[c+d]++; int res=0; for(auto const& v:AB) res += v.second * CD[-v.first]; return res; } };
--------------改进一点儿------------ CD 那里不保存了。而只是-------
class Solution { public: int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) { unordered_map<int,int> AB; for(auto a:A) for(auto b:B) AB[a+b]++; int res = 0; for(auto c: C) for(auto d: D) res += AB[-c -d]; return res; } };
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