Given four lists A, B, C, Dof 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 ofN 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
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class Solution {
public:
intfourSumCount(vector<int>& A, vector<int>& B,vector<int>& C, vector<int>& D) {
vector<int> temp1;
vector<int> temp2;
int count1=0;
for(int i=0;i<A.size(); i++) {
for(intj=0; j<B.size(); j++) {
temp1.push_back(A[i]+B[j]);
}
}
for(int i=0;i<A.size(); i++) {
for(intj=0; j<B.size(); j++) {
temp2.push_back(C[i]+D[j]);
}
}
for(int i=0;i<temp1.size(); i++) {
vector<int>::iterator it;
it=find(temp2.begin(),temp2.end(),0-temp1[i]);
if(it!=temp2.end())
{
intnum = count(temp2.begin(),temp2.end(),0-temp1[i]);
count1 += num;
}
}
returncount1;
}
};