1304: Find-N-Unique-Integers-Sum-up-to-Zero
Easy
table of contents
Simply add all integers between [-n/2, -1] and
[1, n/2] into the ans array as they will
always sum to 0.
Then, if n is odd, we have to add one more unique
integer, 0, into the array, which will not affect the sum
of the pre-existing integers. Finally, we can simply just return the
ans array.
code
class Solution {
public:
vector<int> sumZero(int n) {
vector<int> ans;
for (int i = -n/2; i <= -1; ++i) {
ans.push_back(i);
}
if (n % 2 == 1) {
ans.push_back(0);
}
for (int i = 1; i <= n/2; ++i) {
ans.push_back(i);
}
return ans;
}
};